Stochastic Finance: An Introduction with Market Examples · Stochastic Finance: An Introduction with Market Examples Solutions Manual Chapter 1 Exercise1.1 1.ThepossiblevaluesofRareaandb

Stochastic Finance: An Introduction withMarket Examples

Solutions Manual

Chapter 1

Exercise 1.1

1. The possible values of R are a and b.2. We have

IE∗[R] = aP∗(R = a) + bP∗(R = b)

= ab− rb− a

+ br − ab− a

= r.

3. By Theorem 1.1, there do not exist arbitrage opportunities in this marketsince there exists a risk-neutral measure P∗ from Question 2.

4. The risk-neutral measure is unique hence the market model is completeby Theorem 1.2.

5. Takingη = α(1 + b)− β(1 + a)

π1(b− a) and ξ = β − αS0(b− a) ,

we check that ηπ1 + ξS0(1 + a) = α

ηπ1 + ξS0(1 + b) = β,

which shows thatηπ1 + ξS1 = C.

6. We have

π(C) = ηπ0 + ξS0

= α(1 + b)− β(1 + a)(1 + r)(b− a) + α− β

a− b

= α(1 + b)− β(1 + a)− (1 + r)(α− β)(1 + r)(b− a)

= αb− βa− r(α− β)(1 + r)(b− a) . (S.1.1)

7. We have

1

N. Privault

IE∗[C] = αP∗(R = a) + βP∗(R = b)

= αb− rb− a

+ βr − ab− a

. (S.1.2)

8. Comparing (S.1.1) and (S.1.2) above we do obtain

π(C) = 11 + r

IE∗[C]

9. The initial value π(C) of the portfolio is interpreted as the arbitrage priceof the option contract and it equals the expected value of the discountedpayoff.

10. We have

C = (K − S1)+ = (11− S1)+ =

11− S1 if K > S1,

0 if K ≤ S1.

11. We have

ξ = −(11− (1 + a))b− a

= −23 , η = (1 + b)(11− (1 + a))

(1 + r)(b− a) = 81.05 .

12. The arbitrage price π(C) of the contingent claim C is

π(C) = ηπ0 + ξS0 = 6.952.

Chapter 2

Exercise 2.1

1. The possible values of Rt are a and b.2. We have

IE∗[Rt+1 | Ft] = aP∗(Rt+1 = a | Ft) + bP∗(Rt+1 = b | Ft)

= ab− rb− a

+ br − ab− a

= r.

3. We have

IE∗[St+k | Ft] =k∑i=0

(r − ab− a

)i(b− rb− a

)k−i(k

i

)(1 + b)i(1 + a)k−iSt

= St

k∑i=0

(k

i

)(r − ab− a

(1 + b))i(

b− rb− a

(1 + a))k−i

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Stochastic Finance - Solutions Manual

= St

(r − ab− a

(1 + b) + b− rb− a

(1 + a))k

= (1 + r)kSt.

Assuming that the formula holds for k = 1, its extension to k ≥ 2 can alsobe proved recursively from the “tower property” (16.22) of conditionalexpectations, as follows:

IE∗[St+k | Ft] = IE∗[IE∗[St+k | Ft+k−1] | Ft]= (1 + r) IE∗[St+k−1 | Ft]= (1 + r) IE∗[IE∗[St+k−1 | Ft+k−2] | Ft]= (1 + r)2 IE∗[St+k−2 | Ft]= (1 + r)2 IE∗[IE∗[St+k−2 | Ft+k−3] | Ft]= (1 + r)3 IE∗[St+k−3 | Ft]= · · ·= (1 + r)k−2 IE∗[St+2 | Ft]= (1 + r)k−2 IE∗[IE∗[St+2 | Ft+1] | Ft]= (1 + r)k−1 IE∗[St+1 | Ft]= (1 + r)kSt.

Chapter 3

Exercise 3.1

1. The condition VN = C reads ηNπN + ξN (1 + a)SN−1 = (1 + a)SN−1 −K

ηNπN + ξN (1 + b)SN−1 = (1 + b)SN−1 −K

from which we deduce ξN = 1 and ηN = −K(1 + r)−N/π0.2. We have ηN−1πN−1 + ξN−1(1 + a)SN−1 = ηNπN−1 + ξN (1 + a)SN−1

ηN−1πN−1 + ξN−1(1 + b)SN−1 = ηNπN−1 + ξN (1 + b)SN−1,

which yields ξN−1 = ξN = 1 and ηN−1 = ηN = −K(1 + r)−N/π0.Similarly, solving the self-financing condition

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N. Privault

ηtπt + ξt(1 + a)St = ηt+1πt + ξt+1(1 + a)St

ηtπt + ξt(1 + b)St = ηt+1πt + ξt+1(1 + b)St,

at time t yields ξt = 1 and ηt = −K(1 + r)−N/π0, t = 1, 2, . . . , N .3. We have

πt(C) = Vt = ηtπt+ξtSt = St−K(1+r)−Nπt/π0 = St−K(1+r)−(N−t).

4. For all t = 0, 1, . . . , N we have

(1 + r)−(N−t) IE∗[C | Ft] = (1 + r)−(N−t) IE∗[SN −K | Ft],= (1 + r)−(N−t) IE∗[SN | Ft]− (1 + r)−(N−t) IE∗[K | Ft]= (1 + r)−(N−t)(1 + r)N−tSt −K(1 + r)−(N−t)

= St −K(1 + r)−(N−t)

= Vt = πt(C).

Exercise 3.2

1. This model admits a unique risk-neutral measure P∗ because we havea < r < b. We have

P∗(Rt = a) = b− rb− a

= 0.07− 0.050.07− (−0.02) ,

andP(Rt = b) = r − a

b− a= 0.05− (−0.02)

0.07− (−0.02) ,

t = 1, . . . , N .2. There are no arbitrage opportunities in this model, due to the existence

of a risk-neutral measure.3. This market model is complete because the risk-neutral measure is

unique.4. We have

C = (SN )2,

henceH = (SN )2/(1 + r)N = h(XN ),

withh(x) = x2(1 + r)−N .

Now we haveVt = vt(Xt),

where the function vt(x) is given by

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vt(x) =N−t∑k=0

(N − t)!k!(N − t− k)!

×(r − ab− a

)k (b− rb− a

)N−t−kh

(x

(1 + b

1 + r

)k (1 + a

1 + r

)N−t−k)

= x2(1 + r)−NN−t∑k=0

(N − t)!k!(N − t− k)!

×(r − ab− a

)k (b− rb− a

)N−t−k (1 + b

1 + r

)2k (1 + a

1 + r

)2(N−t−k)

= x2(1 + r)−NN−t∑k=0

(N − t)!k!(N − t− k)!

×(

(r − a)(1 + b)2

(b− a)(1 + r)2

)k ( (b− r)(1 + a)2

(b− a)(1 + r)2

)N−t−k= x2(1 + r)−N

((r − a)(1 + b)2

(b− a)(1 + r)2 + (b− r)(1 + a)2

(b− a)(1 + r)2

)N−t=x2 ((r − a)(1 + b)2 + (b− r)(1 + a)2)N−t

(1 + r)N−2t(b− a)N−t

=x2 ((r − a)(1 + 2b+ b2) + (b− r)(1 + 2a+ a2)

)N−t(1 + r)N−2t(b− a)N−t

=x2 (r(1 + 2b+ b2)− a(1 + 2b+ b2) + b(1 + 2a+ a2)− r(1 + 2a+ a2)

)N−t(1 + r)N−2t(b− a)N−t

= x2 (1 + r(2 + a+ b)− ab)N−t

(1 + r)N−2t .

5. We have

ξ1t =

vt

(1+b1+rXt−1

)− vt

(1+a1+rXt−1

)Xt−1(b− a)/(1 + r)

= Xt−1

(1+b1+r

)2−(

1+a1+r

)2

(b− a)/(1 + r)(1 + r(2 + a+ b)− ab)N−t

(1 + r)N−2t

= St−1(2 + b+ a) (1 + r(2 + a+ b)− ab)N−t

(1 + r)N−t , t = 1, . . . , N,

representing the quantity of the risky asset to be present in the portfolioat time t. On the other hand we have

ξ0t = Vt − ξ1

tXt

X0t

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N. Privault

= Vt − ξ1tXt

π0

= Xt(1 + r(2 + a+ b)− ab)N−tXt −Xt−1(2 + b+ a)/(1 + r)π0(1 + r)N−2t

= St(1 + r(2 + a+ b)− ab)N−tSt − St−1(2 + b+ a)π0(1 + r)N

= −(St−1)2(1 + r(2 + a+ b)− ab)N−t (1 + a)(1 + b)π0(1 + r)N ,

t = 1, . . . , N .6. Let us check that the portfolio is self-financing. We have

ξt+1 · St = ξ0t+1S

0t + ξ1

t+1S1t

= −(St)2(1 + r(2 + a+ b)− ab)N−t−1 (1 + a)(1 + b)π0(1 + r)N S0

t

+(St)2(2 + b+ a) (1 + r(2 + a+ b)− ab)N−t−1

(1 + r)N−t−1

= (St)2 (1 + r(2 + a+ b)− ab)N−t−1

(1 + r)N−t× ((2 + b+ a)(1 + r)− (1 + a)(1 + b))

= (Xt)2(1 + r(2 + a+ b)− ab)N−t 1(1 + r)N−3t

= (1 + r)tVt= ξt · St, t = 1, . . . , N.

Exercise 3.3

1. We have

Vt = ξtSt + ηtπt

= ξt(1 +Rt)St−1 + ηt(1 + r)πt−1.

2. We have

IE∗[Rt|Ft−1] = aP∗(Rt = a | Ft−1) + bP∗(Rt = b | Ft−1)

= ab− rb− a

+ br − ab− a

= br

b− a− a r

b− a= r.

3. By the result of Question 1 we have

IE∗[Vt | Ft−1] = IE∗[ξt(1 +Rt)St−1 | Ft−1] + IE∗[ηt(1 + r)πt−1 | Ft−1]

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= ξtSt−1 IE∗[1 +Rt | Ft−1] + (1 + r) IE∗[ηtπt−1 | Ft−1]= (1 + r)ξtSt−1 + (1 + r)ηtπt−1

= (1 + r)ξtSt + (1 + r)ηtπt= (1 + r)Vt−1,

where we used the self-financing condition.4. We have

Vt−1 = 11 + r

IE∗[Vt | Ft−1]

= 31 + r

P∗(Rt = a | Ft−1) + 81 + r

P∗(Rt = b | Ft−1)

= 11 + 0.15

(30.25− 0.15

0.25− 0.05 + 80.15− 0.050.25− 0.05

)= 1

1.15

(32 + 8

2

)= 4.78.

Chapter 4

Exercise 4.1

1. We need to check whether the four properties of the definition of Brow-nian motion are satisfied. Checking Conditions (i) to (iii) does not poseany particular problem since the time changes t 7→ c + t, t 7→ t/c2 andt 7→ ct2 are deterministic, continuous, and increasing. As for Condi-tion (iv), Bc+t −Bc+s clearly has a centered Gaussian distribution withvariance t, and the same property holds for cBt/c2 since

Var(c(Bt/c2 −Bs/c2)) = c2Var(Bt/c2 −Bs/c2) = c2(t− s)/c2 = t− s.

As a consequence, (a) and (b) are standard Brownian motions.

Concerning (c), we note that Bct2 is a centered Gaussian random variablewith variance ct2 - not t, hence (Bct2)t∈R+ is not a standard Brownianmotion.

2. We havew T

02dBt = 2(BT −B0) = 2BT , which has a Gaussian law with

mean 0 and variance 4T . On the other hand,w T

0(2×1[0,T/2](t)+1(T/2,T ](t))dBt = 2(BT/2−B0)+(BT−BT/2) = BT+BT/2,

which has a Gaussian law with mean 0 and variance 4(T/2)+T/2 = 5T/2.

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N. Privault

3. The stochastic integralw 2π

0sin(t) dBt has a Gaussian distribution with

mean 0 and variancew 2π

0sin2(t)dt =

w 2π

0

1− cos(2t)2 dt = π.

4. If 0 ≤ s ≤ t we have

IE[BtBs] = IE[(Bt−Bs)Bs]+IE[B2s ] = IE[(Bt−Bs)] IE[Bs]+IE[B2

s ] = 0+s = s,

and similarly we obtain IE[BtBs] = t when 0 ≤ t ≤ s, hence in generalwe have IE[BtBs] = min(s, t), s, t ≥ 0.

5. We have

d(f(t)Bt) = f(t)dBt +Btdf(t) + df(t) · dBt= f(t)dBt +Btf

′(t)dt+ f ′(t)dt · dBt= f(t)dBt +Btf

′(t)dt,

and by integration on both sides we get

0 = f(T )BT − f(0)B0

=w T

0d(f(t)Bt)

=w T

0f(t)dBt +

w T

0Btf

′(t)dt,

hence the conclusion.

Exercise 4.2 Let f ∈ L2([0, T ]). We have

E[er T

0 f(s)dBs∣∣∣Ft] = exp

(w t

0f(s)dBs + 1

2

w T

0|f(s)|2ds

), 0 ≤ t ≤ T.

Exercise 4.3 We have

E

[exp

(β

w T

0BtdBt

)]= E

[exp

(β(B2

T − T )/2)]

= e−βT/2E[exp

(eβ(BT )2/2

)]= e−βT/2√

2πT

w∞

−∞e(β− 1

T ) x22 dx

= e−βT/2√

1− βT.

for all β < 1/T .

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July 22, 2018

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Exercise 4.4 We have f(t) = f(0)ect (interest rate compounding) andSt = S0e

σBt−σ2t/2+rt, t ∈ R+, (geometric Brownian motion).

Exercise 4.5

1. By (4.24) we have

d(XTt /(T − t)) = dXT

t

T − t+ XT

t

(T − t)2 dt = σdBtT − t

,

hence by integration using the initial condition X0 = 0 we have

XTt

T − t= σ

w t

0

1T − s

dBs, t ∈ [0, T ].

2. We haveIE[XT

t ] = σ(T − t) IE[w t

0

1T − s

dBs

]= 0.

3. Using the Itô isometry we have

Var[XTt ] = σ2(T − t)2 Var

[w t

0

1T − s

dBs

]= σ2(T − t)2

w t

0

1(T − s)2 ds

= σ2(T − t)2(

1T − t

− 1T

)= σ2(1− t/T ).

4. We have Var[XTT ] = 0 hence XT

T = IE[XTT ] = 0 by Question 2.

Exercise 4.6 Exponential Vasicek model.

1. We have zt = e−atz0 + σw t

0e−a(t−s)dBs.

2. We have yt = e−aty0 + θ

a(1− e−at) + σ

w t

0e−a(t−s)dBs.

3. We have dxt = xt

(θ + σ2

2 − a log xt)dt+ σxtdBt.

4. We have rt = exp(e−at log r0 + θ

a(1− e−at) + σ

w t

0e−a(t−s)dBs

), with

η = θ + σ2/2.5. We have

IE[rt] = exp(e−at log r0 + θ

a(1− e−at) + σ2

4a (1− e−2at)).

6. We have limt→∞

IE[rt] = exp(θ

a+ σ2

4a

).

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N. Privault

Exercise 4.7 Cox-Ingersoll-Ross (CIR) model.

1. We have rt = r0 +w t

0(α− βrs)ds+ σ

w t

0

√rsdBs.

2. Using the fact that the expectation of the stochastic integral with respectto Brownian motion is zero, we get, taking expectations on both sides ofthe above integral equation: u′(t) = α− βu(t).

3. Apply Itô’s formula to

r2t = f

(r0 +

w t

0(α− βrs)ds+ σ

w t

0

√rsdBs

),

with f(x) = x2, to obtain

d(rt)2 = rt(σ2 + 2α− 2βrt)dt+ 2rtσ√rtdBt. (S.4.3)

4. Taking again the expectation on both sides of (S.4.3) we get

IE[r2t ] = IE[r2

0] +w t

0(σ2 IE[rt] + 2α IE[rt]− 2β IE[r2

t ])dt,

and after differentiation with respect to t this yields

v′t = (σ2 + 2α)u(t)− 2βv(t).

Exercise 4.8

1. We have

St = eXt

= eX0 +w t

0use

XsdBs +w t

0vse

Xsds+ 12

w t

0u2seXsds

= eX0 + σw t

0eXsdBs + ν

w t

0eXsds+ σ2

2

w t

0eXsds

= S0 + σw t

0SsdBs + ν

w t

0Ssds+ σ2

2

w t

0Ssds.

2. Let r > 0. The process (St)t∈R+ satisfies the stochastic differential equa-tion

dSt = rStdt+ σStdBt

when r = ν + σ2/2.3. Let the process (St)t∈R+ be defined by St = S0e

σBt+νt, t ∈ R+. Usingthe decomposition ST = Ste

σ(BT−Bt)+ντ , we have

P(ST > K | St = x) = P(Steσ(BT−Bt)+ν(T−t) > K | St = x)= P(xeσ(BT−Bt)+ν(T−t) > K)= P(eσ(BT−Bt) > Ke−ν(T−t)/x)

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July 22, 2018

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= Φ

(− log(Ke−ν(T−t)/x)

σ√τ

)= Φ

(log(x/K) + ντ

σ√τ

),

where τ = T − t.4. We have

η2 = Var[X] = Var[σ(BT −Bt)] = σ2 Var[BT −Bt] = σ2(T − t),

hence η = σ√T − t.

Chapter 5

Exercise 5.1

1. We haveSt = S0e

αt + σw t

0eα(t−s)dBs.

2. We have αM = r.3. After computing the conditional expectation

C(t, x) = e−r(T−t) exp(xer(T−t) + σ2

4r (e2r(T−t) − 1)).

4. Here we need to note that the usual Black-Scholes argument applies andyields ζt = ∂C(t, St)/∂x, that is

ζt = exp(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1)).

Exercise 5.2

1. We have, counting approximately 46 days to maturity,

d− =(r − 1

2σ2)(T − t) + log St

K

σ√T − t

=(0.04377− 1

2 (0.9)2)(46/365) + log 17.236.08

0.9√

46/365= −2.46,

andd+ = d− + 0.9

√46/365 = −2.14.

From the attached table we get

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N. Privault

Φ(d+) = Φ(−2.14) = 0.0162

andΦ(d−) = Φ(−2.46) = 0.0069,

hence

f(t, St) = StΦ(d+)−Ke−r(T−t)Φ(d−)= 17.2× 0.0162− 36.08 ∗ e−0.04377×46/365 × 0.0069= HK$ 0.031.

2. We haveηt = ∂f

∂x(t, St) = Φ(d+) = Φ(−2.14) = 0.0162,

hence one should only hold a fractional quantity 16.2 of the risky assetin order to hedge 1000 such call options when σ = 0.90.

3. From the curve it turns out that when f(t, St) = 10× 0.023 = HK$ 0.23,the volatility σ is approximately equal to σ = 122%.

This approximate value of implied volatility can be found under the col-umn “Implied Volatility (IV.)” on this set of market data from the HongKong Stock Exchange:Derivative Warrant Search

Link to Relevant Exchange Traded Options

Updated: 6 November 2008 Basic Data Market Data

DWCode

Issuer UL Call/Put

DWType

Listing(D-M-Y)

Maturity(D-M-Y)

Strike Entitle-ment

Ratio^

TotalIssueSize

O/S(%)

Delta(%)

IV.(%)

DayHigh($)

DayLow($)

ClosingPrice #

($)

T/O('000)

ULPrice($)

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01897 FB 00066 Call Standard 18-12-2007 23-12-2008 36.08 10 138,000,000 16.43 0.780 125.375 0.000 0.000 0.023 0 17.200

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^ The entitlement ratio in general represents the number of derivative warrants required to be exercised into one share or one unit of the underlying asset (subject to any adjustments as may benecessary to reflect any capitalization, rights issue, distribution or the like).# Closing price value = 99999999 when it is not available.Delayed data on Delta and Implied Volatility of Derivative Warrants are provided by Reuters.Implied Volatility is displayed as 200 even if its value is greater than 200.Users should not use such data provided by Reuters for commercial purposes without its prior written consent.

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Print http://www.hkex.com.hk/dwrc/search/dwlist.asp

1 of 1 11/08/2008 01:57 PM

Derivative Warrant Search Link to Relevant Exchange Traded Options

Updated: 6 November 2008 Basic Data Market Data

DWCode

Issuer UL Call/Put

DWType

Listing(D-M-Y)

Maturity(D-M-Y)

Strike Entitle-ment

Ratio^

TotalIssueSize

O/S(%)

Delta(%)

IV.(%)

DayHigh($)

DayLow($)

ClosingPrice #

($)

T/O('000)

ULPrice($)

Base ListingDocument

Supplemental&

Announcements

01897 FB 00066 Call Standard 18-12-2007 23-12-2008 36.08 10 138,000,000 16.43 0.780 125.375 0.000 0.000 0.023 0 17.200

04348 BP 00066 Call Standard 18-12-2007 23-02-2009 38.88 10 300,000,000 0.25 0.767 88.656 0.000 0.000 0.024 0 17.200

04984 AA 00066 Call Standard 02-06-2005 22-12-2008 12.88 10 300,000,000 0.36 8.075 128.202 0.000 0.000 0.540 0 17.200

05931 SB 00066 Call Standard 27-03-2008 29-12-2008 27.868 10 200,000,000 0.04 2.239 126.132 0.000 0.000 0.086 0 17.200

09133 CT 00066 Call Standard 31-01-2008 08-12-2008 36.88 10 200,000,000 0.15 0.416 133.443 0.000 0.000 0.010 0 17.200

13436 SG 00066 Call Standard 14-05-2008 30-04-2009 32 10 200,000,000 0.10 1.059 61.785 0.000 0.000 0.031 0 17.200

13562 BP 00066 Call Standard 26-05-2008 08-12-2008 30 10 150,000,000 0.00 0.987 127.080 0.000 0.000 0.026 0 17.200

13688 RB 00066 Call Standard 04-06-2008 20-02-2009 26.6 10 200,000,000 7.17 0.706 49.625 0.000 0.000 0.013 0 17.200

13764 SG 00066 Call Standard 13-06-2008 26-02-2009 28 10 300,000,000 0.31 0.549 49.880 0.010 0.010 0.010 6 17.200

13785 ML 00066 Call Standard 17-06-2008 19-01-2009 27.38 10 100,000,000 0.50 0.714 63.598 0.000 0.000 0.014 0 17.200

13821 JP 00066 Call Standard 18-06-2008 18-12-2008 28.8 10 100,000,000 0.81 0.670 91.664 0.000 0.000 0.014 0 17.200

14111 UB 00066 Call Standard 09-07-2008 16-02-2009 23.88 10 500,000,000 0.88 2.288 66.247 0.000 0.000 0.068 0 17.200

14264 BI 00066 Call Standard 16-07-2008 25-02-2009 26.38 10 200,000,000 0.03 1.250 58.172 0.000 0.000 0.030 0 17.200

14305 DB 00066 Call Standard 22-07-2008 09-03-2009 27 10 300,000,000 0.00 0.000 0.000 0.000 0.000 99,999,999.000 0 17.200

14489 FB 00066 Call Standard 06-08-2008 29-06-2009 28.08 10 175,000,000 0.15 1.681 52.209 0.000 0.000 0.053 0 17.200

14512 MB 00066 Call Standard 08-08-2008 26-02-2009 26 10 300,000,000 0.86 1.273 56.527 0.000 0.000 0.030 0 17.200

14531 UB 00066 Call Standard 08-08-2008 11-05-2009 26.88 10 500,000,000 0.00 3.261 84.055 0.000 0.000 0.162 0 17.200

14548 CT 00066 Call Standard 12-08-2008 25-05-2009 26.88 10 400,000,000 0.03 2.749 70.622 0.000 0.000 0.117 0 17.200

14571 CT 00066 Put Standard 15-08-2008 22-06-2009 26 10 240,000,000 0.06 (6.751) 71.243 0.000 0.000 1.020 0 17.200

14925 CT 00066 Call Standard 18-09-2008 12-10-2009 28.88 10 330,000,000 0.00 0.000 0.000 0.000 0.000 99,999,999.000 0 17.200

15159 JP 00066 Call Standard 29-09-2008 12-10-2009 28.08 10 100,000,000 2.00 3.270 66.344 0.000 0.000 0.169 0 17.200

15257 RB 00066 Call Standard 15-10-2008 30-07-2009 22 10 100,000,000 0.41 4.158 60.731 0.280 0.201 0.201 234 17.200

15339 CT 00066 Call Standard 17-10-2008 19-10-2009 20 10 200,000,000 0.07 5.085 58.805 0.290 0.290 0.285 12 17.200

^ The entitlement ratio in general represents the number of derivative warrants required to be exercised into one share or one unit of the underlying asset (subject to any adjustments as may benecessary to reflect any capitalization, rights issue, distribution or the like).# Closing price value = 99999999 when it is not available.Delayed data on Delta and Implied Volatility of Derivative Warrants are provided by Reuters.Implied Volatility is displayed as 200 even if its value is greater than 200.Users should not use such data provided by Reuters for commercial purposes without its prior written consent.

For Underlying Stock price, please refer to Today's Stock Quote of Investment Service Centre.If you are interested in real time data subscription, please click to view the list of Realtime Data Information Vendors.

Print http://www.hkex.com.hk/dwrc/search/dwlist.asp

1 of 1 11/08/2008 01:57 PM

Remark: a typical value for the volatility in standard market conditionswould be around 20%. The observed volatility value σ = 1.22 per year isactually quite high.

Exercise 5.3

1. We find h(x) = x−K.2. Letting g(t, x), the PDE rewrites as

r(x− α(t)) = −α′(t) + rx,

hence α(t) = α(0)ert and g(t, x) = x− α(0)e−rt. The final condition

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July 22, 2018

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g(T, x) = h(x) = x−K

yields α(0) = KerT and g(t, x) = x−Ke−r(T−t).3. We have

ξt = ∂g

∂x(t, St) = 1.

Exercise 5.4

1. We have

Ct = e−r(T−t)E[ST −K | Ft]= e−r(T−t)E[ST | Ft]−Ke−r(T−t)

= ertE[e−rTST | Ft]−Ke−r(T−t)

= erte−rtSt −Ke−r(T−t)

= St −Ke−r(T−t).

We can check that the function g(x, t) = x − Ke−r(T−t) satisfies theBlack-Scholes PDE

rg(x, t) = ∂g

∂t(x, t) + rx

∂g

∂x(x, t) + σ2

2 x2 ∂g

∂x(x, t)

with terminal condition g(x, T ) = x−K, since ∂g(x, t)/∂t = −rKe−r(T−t)

and ∂g(x, t)/∂x = 1.2. We simply take ξt = 1 and ηt = −Ke−rT in order to have

Ct = ξtSt + ηtert = St −Ke−r(T−t), t ∈ [0, T ].

Hint: Find the quantity ξt of the risky asset St and the quantity ηt of theriskless asset ert such that the equality

Ct = ξtSt + ηtert

holds at any time t ∈ [0, T ].

Remark: This hedging strategy is constant over time, and the relationξt = ∂g(St, t)/∂x for the delta is satisfied.

Chapter 6

Exercise 6.1

1. For all t ∈ [0, T ] we have

" 13

N. Privault

C(t, St) = e−r(T−t)S2t IE

[S2T

S2t

]= e−r(T−t)S2

t IE[e2σ(BT−Bt)−σ2(T−t)+2r(T−t)

]= S2

t e(r+σ2)(T−t).

2. For all t ∈ [0, T ] we have

ξt = ∂C

∂x(t, x)|x=St = 2Ste(r+σ2)(T−t),

and

ηt = C(t, St)− ξtStAt

= e−rt

A0(S2t e

(r+σ2)(T−t) − 2S2t e

(r+σ2)(T−t))

= −S2t

A0eσ

2(T−t)+r(T−2t).

3. We have

dC(t, St) = d(S2t e

(r+σ2)(T−t))= −(r + σ2)e(r+σ2)(T−t)S2

t dt+ e(r+σ2)(T−t)d(S2t )

= −(r + σ2)e(r+σ2)(T−t)S2t dt+ e(r+σ2)(T−t)(2StdSt + σ2S2

t dt)= −re(r+σ2)(T−t)S2

t dt+ 2Ste(r+σ2)(T−t)dSt,

and

ξtdSt + ηtdAt = 2Ste(r+σ2)(T−t)dSt − rS2t

A0eσ

2(T−t)+r(T−2t)Atdt

= 2Ste(r+σ2)(T−t)dSt − rS2t eσ2(T−t)+r(T−t)dt,

hence we can check that the strategy is self-financing since dC(t, St) =ξtdSt + ηtdAt.

Exercise 6.2

1. We haveSt = S0e

rt + σw t

0er(t−s)dBs.

2. We haveSt = S0 + σ

w t

0e−rsdBs,

which is a martingale, being a stochastic integral with respect to Brow-nian motion.

This fact can also be proved directly by computing the conditional ex-pectation E[St | Fs] and showing it is equal to Ss:

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E[St | Fs] = E

[S0 + σ

w t

0e−rudBu | Fs

]= E[S0] + σE

[w t

0e−rudBu | Fs

]= S0 + σE

[w s

0e−rudBu | Fs

]+ σE

[w t

se−rudBu | Fs

]= S0 + σ

w s

0e−rudBu + σE

[w t

se−rudBu

]= S0 + σ

w s

0e−rudBu

= Ss.

3. We have

C(t, St) = e−r(T−t)E[exp(ST )|Ft]

= e−r(T−t)E

[exp

(erTS0 + σ

w T

0er(T−u)dBu

) ∣∣∣Ft]= e−r(T−t)E

[exp

(erTS0 + σ

w t

0er(T−u)dBu + σ

w T

ter(T−u)dBu

) ∣∣∣Ft]= exp

(−r(T − t) + er(T−t)St

)E

[exp

(σ

w T

ter(T−u)dBu

) ∣∣∣Ft]= exp


)E

[exp

(σ

w T

ter(T−u)dBu

)]= exp


)exp

(σ2

2

w T

t(er(T−u))2du

)= exp

(−r(T − t) + er(T−t)St + σ2

4r (e2r(T−t) − 1)).

4. We have

ξt = ∂C

∂x(t, St) = exp

(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1))

and

ηt = C(t, St)− ξtStAt

= e−r(T−t)

Atexp

(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1))

− StAt

exp(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1)).

5. We have

" 15

N. Privault

dC(t, St) = re−r(T−t) exp(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1))dt

−rSt exp(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1))dt

−σ2

2 er(T−t) exp(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1))dt

+ exp(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1))dSt

+12e

r(T−t) exp(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1))σ2dt

= re−r(T−t) exp(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1))dt

−rSt exp(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1))dt

+ξtdSt.

On the other hand we have

ξtdSt + ηtdAt = ξtdSt

+re−r(T−t) exp(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1))dt

−rSt exp(Ste

r(T−t) + σ2

4r (e2r(T−t) − 1))dt,

showing thatdC(t, St) = ξtdSt + ηtdAt,

and confirming that the strategy (ξt, ηt)t∈R+ is self-financing.

Exercise 6.3

1. We have

∂f

∂t(t, x) = (r − σ2/2)f(t, x), ∂f

∂x(t, x) = σf(t, x),

and∂2f

∂x2 (t, x) = σ2f(t, x),

hence

dSt = df(t, Bt)

= ∂f

∂t(t, Bt)dt+ ∂f

∂x(t, Bt)dBt + 1

2∂2f

∂x2 (t, Bt)dt

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July 22, 2018

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=(r − 1

2σ2)f(t, Bt)dt+ σf(t, Bt)dBt + 1

2σ2f(t, Bt)dt

= rf(t, Bt)dt+ σf(t, Bt)dBt= rStdt+ σStdBt.

2. We have

E[eσBT |Ft] = E[eσ(BT−Bt+Bt)|Ft]= eσBtE[eσ(BT−Bt)|Ft]= eσBtE[eσ(BT−Bt)]= eσBt+σ

2(T−t)/2.

3. We have

E[ST |Ft] = E[eσBT+rT−σ2T/2|Ft]= erT−σ

2T/2E[eσBT |Ft]= erT−σ

2T/2eσBt+σ2(T−t)/2

= erT+σBt−σ2t/2

= er(T−t)+σBt+rt−σ2t/2

= er(T−t)St.

4. We have

Vt = e−r(T−t)E[C|Ft]= e−r(T−t)E[ST −K|Ft]= e−r(T−t)E[ST |Ft]− e−r(T−t)E[K|Ft]= St − e−r(T−t)K.

5. We take ξt = 1 and ηt = −Ke−rT /A0, t ∈ [0, T ].6. We have

VT = E[C | FT ] = C.

Exercise 6.4 Digital options.

1. By definition of the indicator functions 1[K,∞) and 1[0,K] we have

1[K,∞)(x) =

1 if x ≥ K,

0 if x < K,resp. 1[0,K](x) =

1 if x ≤ K,

0 if x > K,

which shows the claimed result by the definition of Cd and Pd.2. We have

πt(Cd) + πt(Pd) = e−r(T−t) IE[Cd | Ft] + e−r(T−t) IE[Pd | Ft]

" 17

N. Privault

= e−r(T−t) IE[Cd + Pd | Ft]= e−r(T−t) IE[1[K,∞)(ST ) + 1[0,K](ST ) | Ft]= e−r(T−t) IE[1[0,∞)(ST ) | Ft]= e−r(T−t) IE[1 | Ft]= e−r(T−t), 0 ≤ t ≤ T,

since P(ST = K) = 0.3. We have

πt(Cd) = e−r(T−t) IE[Cd | Ft]= e−r(T−t) IE[1[K,∞)(ST ) | St]= e−r(T−t)P (ST ≥ K | St)= Cd(t, St).

4. We have

Cd(t, x) = e−r(T−t)P (ST > K | St = x)

= e−r(T−t)Φ

(rτ − σ2τ/2 + log(x/K)

σ√τ

),

where τ = T − t.5. We have

πt(Cd) = Cd(t, St)

= e−r(T−t)Φ

(rτ − σ2τ/2 + log(St/K)

σ√τ

)= e−r(T−t)Φ (d−) ,

whered− = (r − σ2/2)τ + log(St/K)

σ√τ

.

6. We have

πt(Pd) = e−r(T−t) − πt(Cd)

= e−r(T−t) − e−r(T−t)Φ


σ√τ

)= e−r(T−t)(1− Φ(d−))= e−r(T−t)Φ(−d−).

7. We have

ξt = ∂Cd∂x

(t, St)

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"


= e−r(T−t) ∂

∂xΦ


σ√τ

)x=St

= e−r(T−t) 1σ√

2πτSte−(d−)2/2

> 0.

The Black-Scholes hedging strategy of such a call option does not involveshort-selling because ξt > 0 for all t.

8. Here we have

ξt = ∂Pd∂x

(t, St)

= e−r(T−t) ∂

∂xΦ

(−rτ − σ

2τ/2 + log(x/K)σ√τ

)x=St

= −e−r(T−t) 1σ√

2πτSte−(d−)2/2

< 0.

The Black-Scholes hedging strategy of such a call option does involveshort-selling because ξt < 0 for all t.

Chapter 8

Exercise 8.1

1. We have

P (τa ≥ t) = P (Xt > a) =w∞

aϕXt(x)dx =

√2πt

w∞

ye−x

2/(2t)dx, y > 0.

2. We have

ϕτa(t) = d

dtP (τa ≤ t)

= d

dt

w∞

aϕXt(x)dx

= −12

√2πt−3/2

w∞

ae−x

2/(2t)dx+ 12

√2πt−3/2

w∞

a

x2

te−x

2/(2t)dx

= 12

√2πt−3/2

(−

w∞

ae−x

2/(2t)dx+ ae−a2/(2t) +

w∞

ae−x

2/(2t)dx)

= a√2πt3

e−a2/(2t), t > 0.

3. We have

" 19

N. Privault

E[(τa)−2] = a√2π

w∞

0t−5/2e−a

2/(2t)dt

= 2a√2π

w∞

0x2e−a

2x2/2dx

= 1a2 ,

by the change of variable x = t−1/2, x2 = 1/t, t = x−2, dt = −2x−3dx.

Remark: We have

E[τa] = a√2π

w∞

0t−1/2e−a

2/(2t)dt = +∞.

Exercise 8.2 Barrier options.

1. By (8.30) and (8.18) we find

ξt = ∂g

∂y(t, St) = Φ

(δT−t+

(StK

))− Φ

(δT−t+

(StB

))+K

Be−r(T−t)

(1− 2r

σ2

)(StB

)−2r/σ2 (Φ

(δT−t−

(B2

KSt

))− Φ

(δT−t−

(B

St

)))+ 2rσ2

(StB

)−1−2r/σ2 (Φ

(δT−t+

(B2

KSt

))− Φ

(δT−t+

(B

St

)))− 2σ√

2π(T − t)

(1− K

B

)exp

(−1

2

(δT−t+

(StB

))2),

0 < St ≤ B, 0 ≤ t ≤ T , cf. also Exercise 7.1-(ix) of ? and Figure 8.13.2. We find

P(YT ≤ a & BT ≥ b) = P(BT ≤ 2a− b), a < b < 0,

hence

fYT ,BT (a, b) = dP(YT ≤ a & BT ≤ b)dadb

= −dP(YT ≤ a & BT ≥ b)dadb

, a, b ∈ R.,

satisfies

fYT ,BT (a, b) =√

2πT

1(−∞,b∧0](a) (b− 2a)T

e−(2a−b)2/(2T )

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July 22, 2018

"


=

√

2πT

(b− 2a)T

e−(2a−b)2/(2T ), a < b ∧ 0,

0, a > b ∧ 0.

3. We find

fYT ,BT (a, b) = 1(−∞,b∧0](a) 1T

√2πT

(b− 2a)e−µ2T/2+µb−(2a−b)2/(2T )

=

1T

√2πT

(2a− b)e−µ2T/2+µb−(2a−b)2/(2T ), a < b ∧ 0,

0, a > b ∧ 0.

4. The function g(t, x) is given in Relations (8.12) and (8.13).

Exercise 8.3 Lookback options. By (8.21) and (8.22) we find

ξt = ∂f

∂x(t, St,M t

0)

= −1 +(

1 + 2rσ2

)Φ

(δT−t+

(StM t

0

))+e−r(T−t)

(M t

0St

)2r/σ2 (1− σ2

2r

)Φ

(−δT−t−

(M t

0St

)),

t ∈ [0, T ], and

ηtAt = f(t, St,M t0)− ξtSt

= M t0e−r(T−t)Φ

(−δT−t−

(StM t

0

))− e−r(T−t)

(M t

0St

)−1+2r/σ2

Φ

(−δT−t−

(M t

0St

)).


e−r(T−t) IE[(

1T

w T

0Sudu− κ

)+ ∣∣∣Ft] = e−r(T−t) IE[

1T

w T

0Sudu− κ

∣∣∣Ft]= e−r(T−t) IE

[1T

w T

0Sudu

∣∣∣Ft]− κe−r(T−t)

= e−r(T−t) 1T

IE[w t

0Sudu

∣∣∣Ft]+ e−r(T−t) 1T

IE[w T

tSudu


= e−r(T−t) 1T

w t

0Sudu+ e−r(T−t) 1

TIE[w T

tSudu


" 21

N. Privault

= e−r(T−t) 1T

w t


T

w T

tIE[Su | Ft]du− κe−r(T−t)

= e−r(T−t) 1T

w t


T

w T

tSte

r(u−t)du− κe−r(T−t)

= e−r(T−t) 1T

w t

0Sudu+ e−r(T−t)St

T

w T−t

0erudu− κe−r(T−t)

= e−r(T−t) 1T

w t

0Sudu+ e−r(T−t) St

rT(er(T−t) − 1)− κe−r(T−t)

= e−r(T−t) 1T

w t

0Sudu+ St

1− e−r(T−t)

rT− κe−r(T−t),

t ∈ [0, T ], cf. ? page 361.

We check that the function f(t, x, y) = e−r(T−t)(y/T − κ) + x(1 −e−r(T−t))/(rT ) satisfies the PDE

rf(t, x, y) = ∂f

∂t(t, x, y) + x

∂f

∂y(t, x, y) + rx

∂f

∂x(t, x, y) + 1

2x2σ2 ∂

2f

∂x2 (t, x, y),

t, x > 0, and the boundary conditions f(t, 0, y) = e−r(T−t)(y/T − κ),0 ≤ t ≤ T , y ∈ R+, and f(T, x, y) = y/T − κ, x, y ∈ R+. However, thecondition limy→−∞ f(t, x, y) = 0 is not satisfied because we need to takey > 0 in the above calculation.

Exercise 8.5 The Asian option price can be written as

e−r(T−t) IE∗[(

1T

w T

0Sudu−K

)+ ∣∣∣Ft] = StIE[(UT )+ | Ut

]= Sth(t, Ut) = Stg(t, Zt),

which shows thatg(t, Zt) = h(t, Ut),

and it remains to use the relation

Ut = 1− e−r(T−t)

rT+ e−r(T−t)Zt, t ∈ [0, T ].

Chapter 9

Exercise 9.1 Stopping times.

1. For any t ∈ R+, the question “is τ > t ?” can be answered based on theobservation of the paths of (Bs)0≤s≤t and of the (deterministic) curve(αe−s/2)0≤s≤t up to the time t. Therefore τ is a stopping time.

22

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2. Since τ is a stopping time and (Bt)t∈R+ is a martingale, the stopping timetheorem shows that (eBt∧τ−(t∧τ)/2)t∈R+ is also a martingale and in partic-ular its expectation E[eBt∧τ−(t∧τ)/2] = E[eB0∧τ−(0∧τ)/2] = E[eB0−0/2] =1 is constantly equal to 1 for all t. This shows that

E[eBτ−τ/2] = E[

limt→∞

eBt∧τ−(t∧τ)/2]

= limt→∞

E[eBt∧τ−(t∧τ)/2] = 1.

Next, we note that we have eBτ = αe−τ/2, hence

αE[e−τ ] = E[eBτ−τ/2] = 1,

i.e.αE[e−τ ] = 1/α ≤ 1.

Remark: note that this argument fails when α < 1 because in that caseτ is not a.s. finite.

3. When 0 ≤ t < 1 the question “is ν > t ?” cannot be answered at time twithout waiting to know the value of B1 at time 1. Therefore ν is not astopping time.

Exercise 9.2

1. Letting A0 = 0,

An+1 = An + IE[Mn+1 −Mn | Fn], n ≥ 0,

andNn = Mn −An, n ∈ N, (S.9.4)

we have,

(i) for all n ∈ N,

IE[Nn+1 | Fn] = IE[Mn+1 −An+1 | Fn]= IE[Mn+1 −An − IE[Mn+1 −Mn | Fn] | Fn]= IE[Mn+1 −An | Fn]− IE[IE[Mn+1 −Mn | Fn] | Fn]= IE[Mn+1 −An | Fn]− IE[Mn+1 −Mn | Fn]= − IE[An | Fn] + IE[Mn | Fn]= Mn −An= Nn,

hence (Nn)n∈N is a martingale with respect to (Fn)n∈N.(ii) We have

An+1 −An = IE[Mn+1 −Mn | Fn]= IE[Mn+1 | Fn]− IE[Mn | Fn]

" 23

N. Privault

= IE[Mn+1 | Fn]−Mn ≥ 0, n ∈ N,

since (Mn)n∈N is a submartingale.(iii) By induction we have

An = An−1 + IE[Mn −Mn−1 | Fn−1], n ≥ 1,

which is Fn−1-measurable provided An is Fn−1-measurable, n ≥ 1.(iv) This property is obtained by construction in (S.9.4).

2. For all bounded stopping times σ and τ such that σ ≤ τ a.s., we have

IE[Mσ] = IE[Nσ] + IE[Aσ]≤ IE[Nσ] + IE[Aτ ]= IE[Nτ ] + IE[Aτ ]= IE[Mτ ],

by (9.11), since (Mn)n∈N is a martingale and (An)n∈N is non-decreasing.

Exercise 9.3 American digital options.

1. The optimal strategy is as follows:

(i) if St ≥ K, then exercise immediately.(ii) if St < K, then wait.

2. The optimal strategy is as follows:

(i) if St > K, then wait.(ii) if St ≤ K, exercise immediately.

3. Based on the answers to Question 1 we set

CAmd (t,K) = 1, 0 ≤ t < T,

andCAmd (T, x) = 0, 0 ≤ x < K.

4. Based on the answers to Question 2, we set

PAmd (t,K) = 1, 0 ≤ t < T,

andPAmd (T, x) = 0, x > K.

5. Starting from St ≤ K, the maximum possible payoff is clearly reachedas soon as St hits the level K before the expiration date T , hence thediscounted optimal payoff of the option is e−r(τK−t)1τK<T.

6. From Relation (8.7) we find

24

July 22, 2018

"


P(τa ≤ u) = Φ

(a− µu√

u

)− e2µaΦ

(−a− µu√

u

),

and by differentiation with respect to u this yields the probability densityfunction

fτa(u) = ∂

∂uP(τa ≤ u) = a√

2πu3e−

(a−µu)22u

of the first hitting time of level a by Brownian motion with drift µ. Giventhe relation

Su = Steσ(Bu−Bt)−σ2(u−t)/2+µ(u−t), u ≥ t,

we find that the probability density function of the first hitting time oflevel K after time t by (Su)u∈[t,∞) is given by

u 7−→ a√2π(u− t)3

e−(a−µ(u−t))2

2(u−t) , u ≥ t,

with µ = σ−1(r − σ2/2) and

a = 1σ

log Kx,

given that St = x. Hence for x ∈ (0,K) we have, letting τ = T − t,

CAmd (t, x) = IE[e−r(τK−t)1τK<T | St = x]

=w T

te−r(s−t) a√

2π(s− t)3e−

(a−µ(s−t))22(s−t) ds

=w τ

0e−rs

a√2πs3

e−(a−µs)2

2s ds

=w τ

0

log(K/x)σ√

2πs3exp

(−rs− 1

2σ2s

(−(r − σ2

2

)s+ log K

x

)2)ds

=(K

x

)( rσ2− 1

2 )±( rσ2 + 1

2 )

×w τ

0

log(K/x)σ√

2πs3exp

(− 1

2σ2s

(±(r + σ2

2

)s+ log K

x

)2)ds

= 1√2π

x

K

w∞

y−e−y

2/2dy + 1√2π

(K

x

)2r/σ2 w∞

y+e−y

2/2dy

= x

KΦ

((r + σ2/2)τ + log(x/K)

σ√τ

)+( xK

)−2r/σ2

Φ

(−(r + σ2/2)τ + log(x/K)

σ√τ

), 0 < x < K,

" 25

N. Privault

wherey± = 1

σ√τ

(±(r + σ2

2

)τ + log K

x

),

and used the decomposition

log Kx

= 12

((r + σ2

2

)s+ log K

x

)+ 1

2

(−(r + σ2

2

)s+ log K

x

).

We check thatCAmd (t,K) = Φ(∞) + Φ(−∞) = 1,

and

CAmd (T, x) = x

KΦ (−∞) +

( xK

)−2r/σ2

Φ (−∞) = 0, x < K,

since τ = 0, which is consistent with the answers to Question 3.7. Starting from St ≥ K, the maximum possible payoff is clearly reached

as soon as St hits the level K before the expiration date T , hence thediscounted optimal payoff of the option is e−r(τK−t)1τK<T.

8. Using the notation and answer to Question 6, for x > K we find, lettingτ = T − t,

PAmd (t, x) = IE[e−r(τK−t)1τK<T | St = x]

=w τ

0e−rs

a√2πs3

e−(a−µs)2

2s ds

=w τ

0

log(x/K)σ√

2πs3exp

(−rs− 1

2σ2s

((r − σ2

2

)s+ log x

K

)2)ds

=(K

x

)( rσ2− 1

2 )±( rσ2 + 1

2 )

×w τ

0

log(x/K)σ√

2πs3exp

(− 1

2σ2s

(∓(r + σ2

2

)s+ log x

K

)2)ds

= 1√2π

x

K

w∞

y−e−y

2/2dy + 1√2π

( xK

)2r/σ2 w∞

y+e−y

2/2dy

= x

KΦ

(−(r + σ2/2)τ − log(x/K)

σ√τ

)+( xK

)−2r/σ2

Φ

((r + σ2/2)τ − log(x/K)

σ√τ

), x > K,

withy± = 1

σ√τ

(∓(r + σ2

2

)τ + log x

K

),

We check that

26

July 22, 2018

"


PAmd (t,K) = Φ(−∞) + Φ(∞) = 1,

and

PAmd (T, x) = x

KΦ (−∞) +

( xK

)−2r/σ2

Φ (−∞) = 0, 0 < x < K,

since τ = 0, which is consistent with the answers to Question 3.9. The call-put parity does not hold for American digital options since forx ∈ (0,K) we have

CAmd (t, x) + PAm

d (t, x) = 1 + x

KΦ

((r + σ2/2)τ + log(x/K)

σ√τ

)+( xK

)−2r/σ2

Φ

(−(r + σ2/2)τ + log(x/K)

σ√τ

),

while for x > K we find

CAmd (t, x) + PAm

d (t, x) = 1 + x

KΦ

(−(r + σ2/2)τ − log(x/K)

σ√τ

)+( xK

)−2r/σ2

Φ

((r + σ2/2)τ − log(x/K)

σ√τ

).

Exercise 9.4 American forward contracts. Consider (St)t∈R+ an asset priceprocess given by

dStSt

= rdt+ σdBt,

where (Bt)t∈R+ is a standard Brownian motion.

1. For all stopping times τ such that t ≤ τ ≤ T we have

IE∗[e−r(τ−t)(K − Sτ )

∣∣∣St] = K IE∗[e−r(τ−t)

∣∣∣St]− IE∗[e−r(τ−t)Sτ

∣∣∣St]= e−r(τ−t)K − St,

since τ ∈ [t, T ] is bounded and (e−rtSt)t∈R+ is a martingale, and theabove quantity is clearly maximized by taking τ = t. Hence we have

f(t, St) = supt≤τ≤T

τ stopping time


∣∣∣St] = K − St,

and the optimal strategy is to exercise immediately at time t.2. Similarly we have

IE∗[e−r(τ−t)(Sτ −K)

∣∣∣St] = IE∗[e−r(τ−t)Sτ

∣∣∣St]−K IE∗[e−r(τ−t)

∣∣∣St]= St −K IE∗

[e−r(τ−t)

∣∣∣St] ," 27

N. Privault

since τ ∈ [t, T ] is bounded and (e−rtSt)t∈R+ is a martingale, and theabove quantity is clearly maximized by taking τ = T . Hence we have

f(t, St) = sup t≤τ≤Tτ stopping time


∣∣∣St] = St − e−r(T−t)K,

and the optimal strategy is to exercise at time T .3. Concerning the perpetual American call forward contract, since u 7→e−r(u−t)Su is a martingale, for all stopping times τ we have∗


∣∣∣St] = IE∗[e−r(τ−t)Sτ

∣∣∣St]−K IE∗[e−r(τ−t)

∣∣∣St]≤ St −K IE∗

[e−r(τ−t)

∣∣∣St]≤ St, t ≥ 0.

On the other hand, for all fixed T > 0 we have

IE∗[e−r(T−t)(ST −K)

∣∣∣St] = IE∗[e−r(T−t)ST

∣∣∣St]−K IE∗[e−r(T−t)

∣∣∣St]= St − e−r(T−t)K, t ≥ 0,

hence

supτ≥t

τ stopping time


∣∣∣St] ≥ (St − e−r(T−t)K), T ≥ t,

and letting T →∞ we get

supτ≥t

τ stopping time


∣∣∣St] ≥ limT→∞

(St − e−r(T−t)K

)= St,

hence we have


τ stopping time


∣∣∣St] = St,

and the optimal strategy τ∗ = +∞ is to wait indefinitely.

Concerning the perpetual American put forward contract we have


τ stopping time


∣∣∣St]≤ sup

t≤τ≤Tτ stopping time

IE∗[e−r(τ−t)(K − Sτ )+

∣∣∣St]∗ by Fatou’s Lemma.

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July 22, 2018

"


= fL∗(St).

On the other hand, for τ = τL∗ we have

(K − SτL∗ ) = (K − L∗) = (K − L∗)∗

since 0 < L∗ = 2Kr/(2r + σ2) < K, hence

fL∗(St) = IE∗[e−r(τ−t)(K − SτL∗ )+

∣∣∣St]= IE∗

[e−r(τ−t)(K − SτL∗ )

∣∣∣St]≤ sup

t≤τ≤Tτ stopping time


∣∣∣St]= f(t, St),

which shows thatf(t, St) = fL∗(St),

i.e. the perpetual American put forward contract has same price andexercise strategy as the perpetual American put option.

Exercise 9.51. The option payoff equals (κ− St)p if St ≤ L.2. We have

fL(St) = IE∗[e−r(τL−t)((κ− SτL)+)p

∣∣∣St]= IE∗

[e−r(τL−t)((κ− L)+)p

∣∣∣St]= (κ− L)p IE∗

[e−r(τL−t)

∣∣∣St] .3. We have

fL(x) = IE∗[e−r(τL−t)(κ− SτL)+

∣∣∣St = x]

=

(κ− x)p, 0 < x ≤ L,

(κ− L)p( xL

)−2r/σ2

, x ≥ L.(S.9.5)

4. By differentiating ddx (κ− x)p = −p(κ− x)p−1 we find

f ′L∗(L∗) = − 2rσ2 (κ− L∗)p (L∗)−2r/σ2−1

(L∗)−2r/σ2 = −p(κ− L∗)p−1,

i.e.2rσ2 (κ− L∗) = pL∗,

" 29

N. Privault

orL∗ = 2r

2r + pσ2κ < κ.

5. By (S.9.5) the price can be computed as

f(t, St) = fL∗(St) =

(κ− St)p, 0 < St ≤ L∗,(

pσ2κ

2r + pσ2

)p(2r + pσ2

2rStκ

)−2r/σ2

, St ≥ L∗.

Exercise 9.6

1. The payoff will be κ− (St)p.2. We have

fL(St) = E∗[e−r(τL−t)(κ− (SτL)p)

∣∣∣St]= E∗

[e−r(τL−t)(κ− Lp)

∣∣∣St]= (κ− Lp)E∗

[e−r(τL−t)

∣∣∣St] .3. We have

fL(x) = E∗[e−r(τL−t)(κ− (SτL)p)

∣∣∣St = x]

=

κ− xp, 0 < x ≤ L,

(κ− Lp)( xL

)−2r/σ2

, x ≥ L.

4. We have

f ′L∗(L∗) = − 2rσ2 (κ− (L∗)p) (L∗)−2r/σ2−1

(L∗)−2r/σ2 = −p(L∗)p−1,

i.e.2rσ2 (κ− (L∗)p) = p(L∗)p,

or

L∗ =(

2rκ2r + pσ2

)1/p< (κ)1/p. (S.9.6)

Remark: We may also compute L∗ by maximizing L 7→ fL(x) for all fixedx. The derivative ∂fL(x)/∂L can be computed as

∂fL(x)∂L

= ∂

∂L

((κ− Lp)

(L

x

)2r/σ2)

30

July 22, 2018

"


= −pLp−1(L

x

)2r/σ2

+ 2rσ2L

−1(κ− Lp)(L

x

)2r/σ2

,

and equating ∂fL(x)/∂L to 0 at L = L∗ yields

−p(L∗)p−1 + 2rσ2 (L∗)−1(κ− (L∗)p) = 0,

which recovers (S.9.6).5. The price can be computed as

f(t, St) = fL∗(St) =

κ− (St)p, 0 < St ≤ L∗,

(κ− (L∗)p) (St)−2r/σ2

(L∗)−2r/σ2 , St ≥ L∗

=

κ− (St)p, 0 < St ≤ L∗,

σ2

2r p(St)−2r/σ2

(L∗)p+2r/σ2, St ≥ L∗,

=

κ− (St)p, 0 < St ≤ L∗,

pσ2κ

2r + pσ2

(2r + pσ2

2rSptκ

)−2r/(pσ2)

< κ, St ≥ L∗.

Exercise 9.7

1. We have that

Zt :=(StS0

)λe−(r−a)λt+λσ2t/2−λ2σ2t/2 = eλσBt−λ

2σ2t/2, t ∈ R+,

is a geometric Brownian motion without drift under the risk-neutral prob-ability measure P∗, hence it is a martingale.

2. By the stopping time theorem we have

IE∗[ZτL ] = IE∗[Z0] = 1,

which rewrites as

IE∗[(

SτLS0

)λe−((r−a)λ−λσ2/2+λ2σ2/2)τL

]= 1,

or, given the relation SτL = L,(L

S0

)λIE∗[e−((r−a)λ−λσ2/2+λ2σ2/2)τL

]= 1,

" 31

N. Privault

i.e.

IE∗[e−rτL

]=(S0L

)λ,

provided we choose λ such that

− ((r − a)λ− λσ2/2 + λ2σ2/2) = −r, (S.9.7)

i.e.0 = λ2σ2/2 + λ(r − a− σ2/2)− r.

This equation admits two solutions

λ =−(r − a− σ2/2)±

√(r − a− σ2/2)2 + 4rσ2/2σ2 ,

and we choose the negative solution

λ =−(r − a− σ2/2)−

√(r − a− σ2/2)2 + 4rσ2/2σ2

since S0/L = x/L > 1 and the expectation IE∗[e−rτL ] < 1 is lower than1 as r ≥ 0.

3. Noting that τL = 0 if S0 ≤ L, for all L ∈ (0,K) we have

IE∗[e−rτL(K − SτL)+

∣∣∣S0 = x]

=

K − x, 0 < x ≤ L,

E[e−rτL(K − L)+

∣∣∣S0 = x], x ≥ L.

=

K − x, 0 < x ≤ L,

(K − L)E[e−rτL

∣∣∣S0 = x], x ≥ L.

=

K − x, 0 < x ≤ L,

(K − L)( xL

)−(r−a−σ2/2)−√

(r−a−σ2/2)2+4rσ2/2σ2

, x ≥ L.

4. In order to compute L∗ we observe that, geometrically, the slope of fL(x)at x = L∗ is equal to −1, i.e.

f ′L∗(L∗) = λ(K − L∗) (L∗)λ−1

(L∗)λ = −1,

orλ(K − L∗) = L∗,

32

July 22, 2018

"


orL∗ = λ

λ− 1K < K.

5. For x ≥ L we have

fL∗(x) = (K − L∗)( xL∗

)λ=(K − λ

λ− 1K)(

xλλ−1K

)λ

=(− K

λ− 1

)(x(λ− 1)λK

)λ=(− K

λ− 1

)(x

−λ

)λ(λ− 1−K

)λ=(x

−λ

)λ(λ− 1−K

)λ−1

=( xK

)λ(λ− 1λ

)λK

1− λ. (S.9.8)

6. Let us check that the relation

fL∗(x) ≥ (K − x)+ (S.9.9)

holds. For all x ≤ K we have

fL∗(x)− (K − x) =( xK

)λ(λ− 1λ

)λK

1− λ + x−K

= K

(( xK

)λ(λ− 1λ

)λ 11− λ + x

K− 1).

Hence it suffices to take K = 1 and to show that for all

L∗ = λ

λ− 1 ≤ x ≤ 1

we have

fL∗(x)− (1− x) = xλ

1− λ

(λ− 1λ

)λ+ x− 1

≥ 0.

Equality to 0 holds for x = λ/(λ− 1). By differentiation of this relationwe get

" 33

N. Privault

f ′L∗(x)− (1− x)′ = λxλ−1(λ− 1λ

)λ 11− λ + 1

= xλ−1(λ− 1λ

)λ−1+ 1

≥ 0,

hence the function fL∗(x)− (1− x) is non-decreasing and the inequalityholds throughout the interval [λ/(λ− 1),K].

On the other hand, using (S.9.7) it can be checked by hand that fL∗ givenby (S.9.8) satisfies the equality

(r − a)xf ′L∗(x) + 12σ

2x2f ′′L∗(x) = rfL∗(x) (S.9.10)

for x ≥ L∗ = λ

λ− 1K. In case

0 ≤ x ≤ L∗ = λ

λ− 1K < K,

we havefL∗(x) = K − x = (K − x)+,

hence the relation(rfL∗(x)− (r − a)xf ′L∗(x)− 1

2σ2x2f ′′L∗(x)

)(fL∗(x)− (K − x)+) = 0

always holds. On the other hand, in that case we also have

(r − a)xf ′L∗(x) + 12σ

2x2f ′′L∗(x) = −(r − a)x,

and to conclude we need to show that

(r − a)xf ′L∗(x) + 12σ

2x2f ′′L∗(x) ≤ rfL∗(x) = r(K − x), (S.9.11)

which is true ifax ≤ rK.

Indeed by (S.9.7) we have

(r − a)λ = r + λ(λ− 1)σ2/2≥ r,

hencea

λ

λ− 1 ≤ r,

34

July 22, 2018

"


since λ < 0, which yields

ax ≤ aL∗ ≤ a λ

λ− 1K ≤ rK.

7. By Itô’s formula and the relation

dSt = (r − a)Stdt+ σStdBt

we have

d(fL∗(St)) = −re−rtfL∗(St)dt+ e−rtdfL∗(St)

= −re−rtfL∗(St)dt+ e−rtf ′L∗(St)dSt + 12e−rtσ2S2

t f′′L∗(St)

= e−rt(−rfL∗(St) + (r − a)Stf ′L∗(St) + 1

2σ2S2

t f′′L∗(St)

)dt

+e−rtσStf ′L∗(St)dBt,

and from Equations (S.9.10) and (S.9.11) we have

(r − a)xf ′L∗(x) + 12σ

2x2f ′′L∗(x) ≤ rfL∗(x),

hencet 7→ e−rtfL∗(St)

is a supermartingale.8. By the supermartingale property of

t 7→ e−rtfL∗(St),

for all stopping times τ we have

fL∗(S0) ≥ IE∗[e−rτfL∗(Sτ )

∣∣∣S0

]≥ IE∗

[e−rτ (K − Sτ )+

∣∣∣S0

],

by (S.9.9), hence

fL∗(S0) ≥ supτ stopping time

IE∗[e−rτ (K − Sτ )+

∣∣∣S0

]. (S.9.12)

9. The stopped process

t 7→ e−rt∧τL∗ fL∗(St∧τL∗ )

is a martingale since it has vanishing drift up to time τL∗ by (S.9.10),and it is constant after time τL∗ , hence by the martingale stopping timeTheorem (9.1) we find

" 35

N. Privault

fL∗(S0) = IE∗[e−rτfL∗(SτL∗ )

∣∣∣S0

]= IE∗

[e−rτfL∗(L∗)

∣∣∣S0

]= IE∗

[e−rτ (K − SτL∗ )+

∣∣∣S0

]≤ sup

τ stopping timeIE∗[e−rτ (K − Sτ )+

∣∣∣S0

].

10. By combining the above results and conditioning at time t instead of time0 we deduce that

fL∗(St) = IE∗[e−r(τL∗−t)(K − SτL∗ )+

∣∣∣St]

=

K − St, 0 < St ≤

λ

λ− 1K,(λ− 1−K

)λ−1(−Stλ

)λ, St ≥

λ

λ− 1K,

for all t ∈ R+, where

τL∗ = infu ≥ t : Su ≤ L.

We note that the perpetual put option price does not depend on the valueof t ≥ 0.

Exercise 9.81. By the definition (9.36) of S1(t) and S2(t) we have

Zt = e−rtS2(t)(S1(t)S2(t)

)α= e−rtS1(t)αS2(t)1−α

= S1(0)αS2(0)1−αe(ασ1+(1−α)σ2)Wt−σ22t/2,

which is a martingale when

σ22 = (ασ1 + (1− α)σ2)2,

i.e.ασ1 + (1− α)σ2 = ±σ2,

which yields either α = 0 or

α = 2σ2σ2 − σ1

> 1,

since 0 ≤ σ1 < σ2.2. We have

36

July 22, 2018

"


IE[e−rτL(S1(τL)− S2(τL))+] = IE[e−rτL(LS2(τL)− S2(τL))+]= (L− 1)+ IE[e−rτLS2(τL)]. (S.9.13)

3. Since τL ∧ t is a bounded stopping time we can write

S2(0)(S1(0)S2(0)

)α= IE

[e−r(τL∧t)S2(τL ∧ t)

(S1(τL ∧ t)S2(τL ∧ t)

)α](S.9.14)

= IE[e−rτLS2(τL)

(S1(τL)S2(τL)

)α1τL<t

]+ IE

[e−rtS2(t)

(S1(t)S2(t)

)α1τL>t

]We have

e−rtS2(t)(S1(t)S2(t)

)α1τL>t ≤ e−rtS2(t)Lα1τL>t ≤ e−rtS2(t)Lα,

hence by a uniform integrability argument,

limt→∞

IE[e−rtS2(t)

(S1(t)S2(t)

)α1τL>t

]= 0,

and letting t go to infinity in (S.9.14) shows that

S2(0)(S1(0)S2(0)

)α= IE

[e−rτLS2(τL)

(S1(τL)S2(τL)

)α]= Lα IE

[e−rτLS2(τL)

],

since S1(τL)/S2(τL) = L/L = 1. The conclusion

IE[e−rτL(S1(τL)− S2(τL))+] = (L− 1)+L−αS2(0)(S1(0)S2(0)

)α(S.9.15)

then follows by an application of (S.9.13).4. In order to maximize (S.9.15) as a function of L we consider the derivative

∂

∂L

L− 1Lα

= 1Lα− α(L− 1)L−α−1 = 0,

which vanishes forL∗ = α

α− 1 ,

and we substitute L in (S.9.15) with the value of L∗.5. In addition to r = σ2

2/2 it is sufficient to let S1(0) = κ and σ1 = 0 whichyields α = 2, L∗ = 2, and we find

supτ stopping time

IE[e−rτ (κ− S2(τ))+] = 1S2(0)

(κ2

)2,

which coincides with the result of Proposition 9.4.

" 37

N. Privault

Chapter 10

Exercise 10.1

1. We have

dXt = d

(Xt

Nt

)= X0N0

d(e(σ−η)Bt−(σ2−η2)t/2

)= X0N0

(σ − η)e(σ−η)Bt−(σ2−η2)t/2dBt + X02N0

(σ − η)2e(σ−η)Bt−(σ2−η2)t/2dt

− X02N0

(σ2 − η2)e(σ−η)Bt−(σ2−η2)t/2dt

= − Xt

2Nt(σ2 − η2)dt+ Xt

Nt(σ − η)dBt + Xt

2Nt(σ − η)2dt

= −Xt

Ntη(σ − η)dt+ Xt

Nt(σ − η)dBt

= Xt

Nt(σ − η)(dBt − ηdt)

= (σ − η)Xt

NtdBt = (σ − η)XtdBt,

where dBt = dBt − ηdt is a standard Brownian motion under P.2. By the result of Question 1, Xt is a driftless geometric Brownian motion

with volatility σ − η under P, hence

IE[(XT −λ)+] = X0Φ

(log(X0/λ)σ√T

+ σ√T

2

)−λΦ

(log(X0/λ)σ√T

− σ√T

2

)

is given by the Black-Scholes formula with zero interest rate and volatilityparameter σ = σ − η, which shows (10.30) by multiplication by N0 andthe relation X0 = N0X0.

Hint 1: We have the change of numéraire identity IE[(XT − λNT )+] =N0IE[(XT − λ)+].

3. We have σ = σ − η.

Exercise 10.2 Bond options.

1. Itô’s formula yields

d

(P (t, S)P (t, T )

)= P (t, S)P (t, T ) (ζS(t)− ζT (t))(dWt − ζT (t)dt)

= P (t, S)P (t, T ) (ζS(t)− ζT (t))dWt, (S.10.16)

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July 22, 2018

"


where (Wt)t∈R+ is a standard Brownian motion under P by the Girsanovtheorem.

2. From (S.10.16) we have

P (t, S)P (t, T ) = P (0, S)

P (0, T ) exp(w t

0(ζS(s)− ζT (s))dWs −

12

w t

0|ζS(s)− ζT (s)|2ds

),

hence

P (u, S)P (u, T ) = P (t, S)

P (t, T ) exp(w u

t(ζS(s)− ζT (s))dWs −

12

w u

t|ζS(s)− ζT (s)|2ds

),

u ≥ t, and for u = T this yields

P (T, S) = P (t, S)P (t, T ) exp

(w T

t(ζS(s)− ζT (s))dWs −

12

w T


),

since P (T, T ) = 1. Let P denote the forward measure associated to thenuméraire

Nt := P (t, T ), 0 ≤ t ≤ T.

3. For all S ≥ T > 0 we have

IE[e−

r Ttrsds(P (T, S)−K)+

∣∣∣Ft]= P (t, T )IE

[(P (t, S)P (t, T ) exp

(X − 1

2

w T


)−K

)+ ∣∣∣Ft]

= P (t, T )IE[(eX+m(t,T,S) −K

)+ ∣∣∣Ft] ,where X is a centered Gaussian random variable with variance

v2(t, T, S) =w T


given Ft, and

m(t, T, S) = −12v

2(t, T, S) + log P (t, S)P (t, T ) .

Recall that when X is a centered Gaussian random variable with variancev2, the expectation of (em+X −K)+ is given, as in the standard Black-Scholes formula, by

IE[(em+X −K)+] = em+ v22 Φ(v + (m− logK)/v)−KΦ((m− logK)/v),

where

" 39

N. Privault

Φ(z) =w z

−∞e−y

2/2 dy√2π, z ∈ R,

denotes the Gaussian cumulative distribution function and for simplic-ity of notation we dropped the indices t, T, S in m(t, T, S) and v2(t, T, S).

Consequently we have

IE[e−

r Ttrsds(P (T, S)−K)+

∣∣∣Ft]= P (t, S)Φ

(v

2 + 1v

log P (t, S)KP (t, T )

)−KP (t, T )Φ

(−v2 + 1

vlog P (t, S)

KP (t, T )

).

4. The self-financing hedging strategy that hedges the bond option is ob-tained by holding a (possibly fractional) quantity

Φ

(v

2 + 1v

log P (t, S)KP (t, T )

)of the bond with maturity S, and by shorting a quantity

KΦ

(−v2 + 1

vlog P (t, S)

KP (t, T )

)of the bond with maturity T .

Exercise 10.3

1. The processe−rtS2(t) = S2(0)eσ2Wt+(µ−r)t

is a martingale ifr − µ = 1

2σ22 .

2. We note that

e−rtXt = e−rte(r−µ)t−σ21t/2S1(t)

= e−rte(σ22−σ

21)t/2S1(t)

= e−µt−σ21t/2S1(t)

= S1(0)eµt−σ21t/2eσ1Wt+µt

= S1(0)eσ1Wt−σ21t/2

is a martingale, where

Xt = e(r−µ)t−σ21t/2S1(t) = e(σ2

2−σ21)t/2S1(t).

3. By (10.32) we have

40

July 22, 2018

"


X(t) = Xt

Nt

= e(σ22−σ

21)t/2S1(t)

S2(t)

= S1(0)S2(0)e

(σ22−σ

21)t/2+(σ1−σ2)Wt

= S1(0)S2(0)e

(σ22−σ

21)t/2+(σ1−σ2)Wt+σ2(σ1−σ2)t

= S1(0)S2(0)e

(σ1−σ2)Wt+σ2σ1t−(σ22+σ2

1)t/2

= S1(0)S2(0)e

(σ1−σ2)Wt−(σ1−σ2)2t/2,

whereWt := Wt − σ2t

is a standard Brownian motion under the forward measure P defined by

dPdP

= e−r T

0 rsdsNTN0

= e−rTS2(T )S2(0)

= e−rT eσ2WT+µT

= eσ2WT+(µ−r)T

= eσ2WT−σ22t/2.

4. Given that Xt = e(σ22−σ

21)t/2S1(t) and X(t) = Xt/Nt = Xt/S2(t), we

have

e−rT IE[(S1(T )− κS2(T ))+] = e−rT IE[(e−(σ22−σ

21)T/2XT − κS2(T ))+]

= e−rT e−(σ22−σ

21)T/2 IE[(XT − κe(σ2

2−σ21)T/2S2(T ))+]

= S2(0)e−(σ22−σ

21)T/2IE[(XT − κe(σ2

2−σ21)T/2)+]

= S2(0)e−(σ22−σ

21)T/2IE[(X0e

(σ1−σ2)WT−(σ1−σ2)2T/2 − κe(σ22−σ

21)T/2)+]

= S2(0)e−(σ22−σ

21)T/2

(X0Φ

0+(T, X0)− κe(σ2

2−σ21)T/2Φ0

−(T, X0))

= S2(0)e−(σ22−σ

21)T/2X0Φ

0+(T, X0)

−κS2(0)e−(σ22−σ

21)T/2e(σ2

2−σ21)T/2Φ0

−(T, X0)

= e−(σ22−σ

21)T/2X0Φ

0+(T, X0)− κS2(0)Φ0

−(T, X0)= S1(0)Φ0

+(T, X0)− κS2(0)Φ0−(T, X0),

where

" 41

N. Privault

Φ0+(T, x) = Φ

(log(x/κ)|σ1 − σ2|

√T

+ (σ1 − σ2)2 − (σ22 − σ2

1)2|σ1 − σ2|

√T

)

=

Φ

(log(x/κ)|σ1 − σ2|

√T

+ σ1√T

), σ1 > σ2,

Φ

(log(x/κ)|σ1 − σ2|

√T− σ1

√T

), σ1 < σ2,

and

Φ0−(T, x) = Φ

(log(x/κ)|σ1 − σ2|

√T− (σ1 − σ2)2 + (σ2

2 − σ21)

2|σ1 − σ2|√T

)

=

Φ

(log(x/κ)|σ1 − σ2|

√T

+ σ2√T

), σ1 > σ2,

Φ

(log(x/κ)|σ1 − σ2|

√T− σ2

√T

), σ1 < σ2,

if σ1 6= σ2. In case σ1 = σ2 we find

e−rT IE[(S1(T )− κS2(T ))+] = e−rT IE[S1(T )(1− κS2(0)/S1(0))+]= (1− κS2(0)/S1(0))+e−rT IE[S1(T )]= (S1(0)− κS2(0))1S1(0)>κS2(0).

Exercise 10.4

1. It suffices to check that the definition of (WNt )t∈R+ implies the correlation

identity dWSt · dWN

t = ρdt by Itô’s calculus.2. We let

σt =√

(σSt )2 − 2ρσRt σSt + (σRt )2

and

dWXt = σSt − ρσNt

σtdWS

t −√

1− ρ2σNt

σtdWt, t ∈ R+,

which defines a standard Brownian motion under P∗ due to the definitionof σt.

Exercise 10.5

1. We have σ =√

(σS)2 − 2ρσRσS + (σR)2.2. Letting Xt = e−rtXt = e(a−r)tSt/Rt, t ∈ R+, we have

IE∗[(

STRT− κ)+ ∣∣∣Ft] = e−aT IE∗

[(XT − eaTκ

)+∣∣∣Ft]

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"


= e−(a−r)T IE∗[(XT − e(a−r)Tκ

)+ ∣∣∣Ft]= e−(a−r)T

(XtΦ

((r − a+ σ2/2)(T − t)

σ√T − t

+ 1σ√T − t

log StκRt

)−κe(a−r)TΦ

((r − a− σ2/2)(T − t)

σ√T − t

+ 1σ√T − t

log StκRt

))= StRte(r−a)(T−t)Φ

((r − a+ σ2/2)(T − t)

σ√T − t

+ 1σ√T − t

log StκRt

)−κΦ

((r − a− σ2/2)(T − t)

σ√T − t

+ 1σ√T − t

log StκRt

),

hence the price of the quanto option is

e−r(T−t) IE∗[(

STRT− κ)+ ∣∣∣Ft]

= StRte−a(T−t)Φ

((r − a+ σ2/2)(T − t)

σ√T − t

+ 1σ√T − t

log StκRt

)−κe−r(T−t)Φ

((r − a− σ2/2)(T − t)

σ√T − t

+ 1σ√T − t

log StκRt

).

Chapter 11

Exercise 11.1

1. We have rt = r0 + at+Bt, and

F (t, rt) = F (t, r0 + at+ σBt),

hence by Proposition 11.2 the PDE satisfied by F (t, x) is

− xF (t, x) + ∂F

∂t(t, x) + a

∂F

∂x(t, x) + 1

2σ2 ∂

2F

∂x2 (t, x) = 0, (S.11.17)

with terminal condition F (T, x) = 1.2. We have rt = r0 + at+Bt and

F (t, rt) = IE∗[exp

(−

w T

trsds

) ∣∣∣Ft]= IE∗

[exp

(−r0(T − t)− a

w T

tsds−

w T

tBsds

) ∣∣∣Ft]= IE∗

[e−r0(T−t)−a(T 2−t2)/2 exp

(−(T − t)Bt −

w T

t(T − s)dBs

) ∣∣∣Ft]" 43

N. Privault

= e−r0(T−t)−a(T 2−t2)/2−(T−t)Bt IE∗[exp

(−

w T

t(T − s)dBs

) ∣∣∣Ft]= e−r0(T−t)−a(T−t)(T+t)/2−(T−t)Bt IE∗

[exp

(−

w T

t(T − s)dBs

)]= exp

(−(T − t)rt − a(T − t)2/2 + 1

2

w T

t(T − s)2ds

)= exp

(−(T − t)rt − a(T − t)2/2 + (T − t)3/6

),

hence F (t, x) = exp(−(T − t)x− a(T − t)2/2 + (T − t)3/6

).

Note that the PDE (S.11.17) can also be solved by looking for a solutionof the form F (t, x) = eA(T−t)+xC(T−t), in which case one would findA(s) = −as2/2 + s3/6 and C(s) = −s.

3. We check that the function F (t, x) of Question 2 satisfies the PDE(S.11.17) of Question 1, since F (T, x) = 1 and

−xF (t, x) +(x+ a(T − t)− (T − t)2

2

)F (t, x)− a(T − t)F (t, x)

+12σ

2(T − t)2F (t, x) = 0.

4. We have

f(t, T, S) = 1S − T

(logP (t, T )− logP (t, S))

= 1S − T

((−(T − t)rt + σ2

6 (T − t)3)−(−(S − t)rt + σ2

6 (S − t)3))

= rt + 1S − T

σ2

6 ((T − t)3 − (S − t)3).

5. We havef(t, T ) = − ∂

∂TlogP (t, T ) = rt −

σ2

2 (T − t)2.

6. We havedtf(t, T ) = σ2(T − t)dt+ σdWt.

7. The HJM condition (11.33) is satisfied since the drift of dtf(t, T ) equalsσ

r Ttσds.

Exercise 11.2

1. We have

P (t, T ) = P (s, T ) exp(w t

srudu+

w t

sσTu dBu −

12

w t

s|σTu |2du

),

0 ≤ s ≤ t ≤ T .

44

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2. We haved(e−

r t0 rsdsP (t, T )

)= e−

r t0 rsdsσTt P (t, T )dBt,

which gives a martingale after integration, from the properties of the Itôintegral.

3. By the martingale property of the previous question we have

IE[e−

r T0 rsds

∣∣∣Ft] = IE[P (T, T )e−

r T0 rsds

∣∣∣Ft]= P (t, T )e−

r t0 rsds, 0 ≤ t ≤ T.

4. By the previous question we have

P (t, T ) = er t

0 rsds IE[e−

r T0 rsds

∣∣∣Ft]= IE

[er t

0 rsdse−r T

0 rsds∣∣∣Ft]

= IE[e−

r Ttrsds

∣∣∣Ft] , 0 ≤ t ≤ T,

since e−r t

0 rsds is an Ft-measurable random variable.5. We have

P (t, S)P (t, T ) = P (s, S)

P (s, T ) exp(w t

s(σSu − σTu )dBu −

12

w t

s(|σSu |2 − |σTu |2)du

)= P (s, S)P (s, T ) exp

(w t

s(σSu − σTu )dBTu −

12

w t

s(σSu − σTu )2du

),

0 ≤ t ≤ T , hence letting s = t and t = T in the above expression we have

P (T, S) = P (t, S)P (t, T ) exp

(w T

t(σSs − σTs )dBTs −

12

w T

t(σSs − σTs )2ds

).

6. We have

P (t, T ) IET[(P (T, S)− κ)+

]= P (t, T ) IET

[(P (t, S)P (t, T )e

r Tt

(σSs −σTs )dBTs − 1

2r Tt

(σSs −σTs )2ds − κ

)+]

= P (t, T ) IE[(eX − κ)+ | Ft]

= P (t, T )emt+v2t /2Φ

(vt2 + 1

vt(mt + v2

t /2− log κ))

−κP (t, T )Φ(−vt2 + 1

vt(mt + v2

t /2− log κ)),

with

" 45

N. Privault

mt = log(P (t, S)/P (t, T ))− 12

w T

t(σSs − σTs )2ds

andv2t =

w T

t(σSs − σTs )2ds,

i.e.

P (t, T ) IET[(P (T, S)− κ)+

]= P (t, S)Φ

(vt2 + 1

vtlog P (t, S)

κP (t, T )

)− κP (t, T )Φ

(−vt2 + 1

vtlog P (t, S)

κP (t, T )

).

Exercise 11.31. We check that P (T, T ) = eX

TT = 1.

2. We have

f(t, T, S) = − 1S − T

(XSt −XT

t − µ(S − T ))

= µ− σ 1S − T

((S − t)

w t

0

1S − s

dBs − (T − t)w t

0

1T − s

dBs

)= µ− σ 1

S − T

w t

0

(S − tS − s

− T − tT − s

)dBs

= µ− σ 1S − T

w t

0

(T − s)(S − t)− (T − t)(S − s)(S − s)(T − s) dBs

= µ+ σ

S − T

w t

0

(s− t)(S − T )(S − s)(T − s)dBs.

3. We have

f(t, T ) = µ− σw t

0

t− s(T − s)2 dBs.

4. We note thatlimTt

f(t, T ) = µ− σw t

0

1t− s

dBs

does not exist in L2(Ω).5. By Itô’s calculus we have

dP (t, T )P (t, T ) = σdBt + 1

2σ2dt+ µdt− XT

t

T − tdt

= σdBt + 12σ

2dt− logP (t, T )T − t

dt, t ∈ [0, T ].

6. Let

rSt = µ+ 1

2σ2 − XS

t

S − t46

July 22, 2018

"


= µ+ 12σ

2 − σw t

0

1S − s

dBs,

and apply the result of Exercise 11.11.5-(4).7. We have

IE[dPTdP

∣∣∣Ft] = eσBt−σ2t/2.

8. By the Girsanov theorem, the process Bt := Bt− σt is a standard Brow-nian motion under PT .

9. We have

logP (T, S) = −µ(S − T ) + σ(S − T )w T

0

1S − s

dBs

= −µ(S − T ) + σ(S − T )w t

0

1S − s

dBs + σ(S − T )w T

t

1S − s

dBs

= S − TS − t

logP (t, S) + σ(S − T )w T

t

1S − s

dBs

= S − TS − t


t

1S − s

dBs + σ2(S − T )w T

t

1S − s

ds

= S − TS − t


t

1S − s

dBs + σ2(S − T ) log S − tS − T

,

0 < T < S.10. We have

P (t, T ) IET[(P (T, S)−K)+

∣∣∣Ft]= P (t, T ) IE[(eX − κ)+ | Ft]


(vt2 + 1

vt(mt + v2

t /2− log κ))

−κP (t, T )Φ(−vt2 + 1

vt(mt + v2

t /2− log κ))


(vt + 1

vt(mt − log κ)

)− κP (t, T )Φ

(1vt

(mt − log κ)),

withmt = S − T

S − tlogP (t, S) + σ2(S − T ) log S − t

S − Tand

v2t = σ2(S − T )2

w T

t

1(S − s)2 ds

= σ2(S − T )2(

1S − T

− 1S − t

)

" 47

N. Privault

= σ2(S − T ) (T − t)(S − t) ,

hence

P (t, T ) IET[(P (T, S)−K)+

∣∣∣Ft]= P (t, T ) (P (t, S))(S−T )(S−t)

(S − tS − T

)σ2(S−T )ev

2t /2

×Φ

(vt + 1

vtlog(

(P (t, S))(S−T )(S−t)

κ

(S − tS − T

)σ2(S−T )))

−κP (t, T )Φ(

1vt

log(

(P (t, S))(S−T )(S−t)

κ

(S − tS − T

)σ2(S−T )))

.

Exercise 11.4 From Proposition 11.2 the bond pricing PDE is∂F

∂t(t, x) = xF (t, x)− (α− βx)∂F

∂x(t, x)− 1

2σ2x2 ∂

2F

∂x2 (t, x)

F (T, x) = 1.

Let us search for a solution of the form

F (t, x) = eA(T−t)−xB(T−t),

with A(0) = B(0) = 0, which impliesA′(s) = 0

B′(s) + βB(s) + 12σ

2B2(s) = 1.

hence in particular A(s) = 0, s ∈ R, and B(s) solves a Riccatti equation,whose solution is easily checked to be

B(s) = 2(eγs − 1)2γ + (β + γ)(eγs − 1) ,

with γ =√β2 + 2σ2.

Chapter 12

Exercise 12.1

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1. The forward measure PS is defined from the numéraire Nt := P (t, S) andthis gives

Ft = P (t, S)IE[(κ− L(T, T, S))+ | Ft].

2. The LIBOR rate L(t, T, S) is a driftless geometric Brownian motion withvolatility σ under the forward measure PS . Indeed, the LIBOR rateL(t, T, S) can be written as the forward price L(t, T, S) = Xt = Xt/Ntwhere Xt = (P (t, T )−P (t, S))/(S−T ) and Nt = P (t, S). Since both dis-counted bond prices e−

r t0 rsdsP (t, T ) and e−

r t0 rsdsP (t, S) are martingales

under P∗, the same is true of Xt. Hence L(t, T, S) = Xt/Nt becomes amartingale under the forward measure PS by Proposition 2.1, and com-puting its dynamics under PS amounts to removing any “dt” term in(12.19), i.e.

dL(t, T, S) = σL(t, T, S)dWt, 0 ≤ t ≤ T,

hence L(t, T, S) = L(0, T, S)eσWt−σ2t/2, where (Wt)t∈R+ is a standardBrownian motion under PS .

3. We find

Ft = P (t, S)IE[(κ− L(T, T, S))+ | Ft]= P (t, S)IE[(κ− L(t, T, S)e−σ

2(T−t)/2+σ(WT−Wt))+ | Ft]= P (t, S)(κΦ(−d−)− XtΦ(−d+))= κP (t, S)Φ(−d−)− P (t, S)L(t, T, S)Φ(−d+)= κP (t, S)Φ(−d−)− (P (t, T )− P (t, S))Φ(−d+)/(S − T ),

where em = L(t, T, S)e−σ2(T−t)/2, v2 = (T − t)σ2, and

d+ = log(L(t, T, S)/κ)σ√T − t

+ σ√T − t2 ,

andd− = log(L(t, T, S)/κ)

σ√T − t

− σ√T − t2 ,

because L(t, T, S) is a driftless geometric Brownian motion with volatilityσ under the forward measure PS .

Exercise 12.2

1. We havedP (t, Ti)P (t, Ti)

= rtdt+ ζitdBt, i = 1, 2,

and

P (T, Ti) = P (t, Ti) exp(w T

trsds+

w T

tζisdBs −

12

w T

t(ζis)2ds

),

" 49

N. Privault

0 ≤ t ≤ T ≤ Ti, i = 1, 2, hence

logP (T, Ti) = logP (t, Ti) +w T

trsds+

w T

tζisdBs −

12

w T

t(ζis)2ds,

0 ≤ t ≤ T ≤ Ti, i = 1, 2, and

d logP (t, Ti) = rtdt+ ζitdBt −12(ζit)2dt, i = 1, 2.

In the present modeldrt = σdBt,

where (Bt)t∈R+ is a standard Brownian motion under P, we have

ζit = −σ(Ti − t), 0 ≤ t ≤ Ti, i = 1, 2.

LettingdBit = dBt − ζitdt,

defines a standard Brownian motion under Pi, i = 1, 2, and we have

P (T, T1)P (T, T2) = P (t, T1)

P (t, T2) exp(w T

t(ζ1s − ζ2

s )dBs −12

w T

t((ζ1

s )2 − (ζ2s )2)ds

)= P (t, T1)P (t, T2) exp

(w T

t(ζ1s − ζ2

s )dB2s −

12

w T

t(ζ1s − ζ2

s )2ds

),

which is an Ft-martingale under P2 and under P1,2, and

P (T, T2)P (T, T1) = P (t, T2)

P (t, T1) exp(−

w T

t(ζ1s − ζ2

s )dB1s −

12

w T

t(ζ1s − ζ2

s )2ds

),

which is an Ft-martingale under P1.2. We have

f(t, T1, T2) = − 1T2 − T1

(logP (t, T2)− logP (t, T1))

= rt + 1T2 − T1

σ2

6 ((T1 − t)3 − (T2 − t)3).

3. We have

df(t, T1, T2) = − 1T2 − T1

d log (P (t, T2)/P (t, T1))

= − 1T2 − T1

((ζ2t − ζ1

t )dBt −12((ζ2

t )2 − (ζ1t )2)dt

)= − 1

T2 − T1

((ζ2t − ζ1

t )(dB2t + ζ2

t dt)−12((ζ2

t )2 − (ζ1t )2)dt

)

50

July 22, 2018

"


= − 1T2 − T1

((ζ2t − ζ1

t )dB2t −

12(ζ2

t − ζ1t )2dt

).

4. We have

f(T, T1, T2) = − 1T2 − T1

log (P (T, T2)/P (T, T1))

= f(t, T1, T2)− 1T2 − T1

(w T

t(ζ2s − ζ1

s )dBs −12((ζ2

s )2 − (ζ1s )2)ds

)= f(t, T1, T2)− 1

T2 − T1

(w T

t(ζ2s − ζ1

s )dB2s −

12

w T

t(ζ2s − ζ1

s )2ds

)= f(t, T1, T2)− 1

T2 − T1

(w T

t(ζ2s − ζ1

s )dB1s + 1

2

w T

t(ζ2s − ζ1

s )2ds

).

Hence f(T, T1, T2) has a Gaussian distribution given Ft with conditionalmean

m = f(t, T1, T2) + 12

w T

t(ζ2s − ζ1

s )2ds

under P2, resp.

m = f(t, T1, T2)− 12

w T

t(ζ2s − ζ1

s )2ds

under P1, and variance

v2 = 1(T2 − T1)2

w T

t(ζ2s − ζ1

s )2ds.

Hence

(T2 − T1) IE[e−

r T2t rsds(f(T1, T1, T2)− κ)+

∣∣∣Ft]= (T2 − T1)P (t, T2) IE2

[(f(T1, T1, T2)− κ)+

∣∣∣Ft]= (T2 − T1)P (t, T2) IE2

[(m+X − κ)+

∣∣∣Ft]= (T2 − T1)P (t, T2)

(v√2πe−

(κ−m)2

2v2 + (m− κ)Φ((m− κ)/v)).

5. We have

L(T, T1, T2) = S(T, T1, T2)

= 1T2 − T1

(P (T, T1)P (T, T2) − 1

)= 1T2 − T1

(P (t, T1)P (t, T2) exp

(w T

t(ζ1s − ζ2

s )dBs −12

w T

t((ζ1

s )2 − (ζ2s )2)ds

)− 1)

" 51

N. Privault

= 1T2 − T1

(P (t, T1)P (t, T2) exp

(w T

t(ζ1s − ζ2

s )dB2s −

12

w T

t(ζ1s − ζ2

s )2ds

)− 1)

= 1T2 − T1

(P (t, T1)P (t, T2) exp

(w T

t(ζ1s − ζ2

s )dB1s + 1

2

w T

t(ζ1s − ζ2

s )2ds

)− 1),

and by Itô calculus,

dS(t, T1, T2) = 1T2 − T1

d

(P (t, T1)P (t, T2)

)= 1T2 − T1

P (t, T1)P (t, T2)

((ζ1t − ζ2

t )dBt + 12(ζ1

t − ζ2t )2dt− 1

2((ζ1t )2 − (ζ2

t )2)dt)

=(

1T2 − T1

+ S(t, T1, T2))(

(ζ1t − ζ2

t )dBt + ζ2t (ζ2

t − ζ1t )dt)dt

)=(

1T2 − T1

+ S(t, T1, T2))(

(ζ1t − ζ2

t )dB1t + ((ζ2

t )2 − (ζ1t )2)dt

)=(

1T2 − T1

+ S(t, T1, T2))

(ζ1t − ζ2

t )dB2t , t ∈ [0, T1],

hence 1T2−T1

+ S(t, T1, T2) is a geometric Brownian motion, with

1T2 − T1

+ S(T, T1, T2)

=(

1T2 − T1

+ S(t, T1, T2))

exp(w T

t(ζ1s − ζ2

s )dB2s −

12

w T

t(ζ1s − ζ2

s )2ds

),

0 ≤ t ≤ T ≤ T1.6. We have

(T2 − T1) IE[e−

r T2t rsds(L(T1, T1, T2)− κ)+

∣∣∣Ft]= (T2 − T1) IE

[e−

r T1t rsdsP (T1, T2)(L(T1, T1, T2)− κ)+

∣∣∣Ft]= P (t, T1, T2) IE1,2

[(S(T1, T1, T2)− κ)+

∣∣∣Ft] .The forward measure P2 is defined by

IE[dP2dP

∣∣∣Ft] = P (t, T2)P (0, T2)e

−r t

0 rsds, 0 ≤ t ≤ T2,

and the forward swap measure is defined by

IE[dP1,2dP

∣∣∣Ft] = P (t, T2)P (0, T2)e

−r t

0 rsds, 0 ≤ t ≤ T1,

52

July 22, 2018

"


hence P2 and P1,2 coincide up to time T1 and (B2t )t∈[0,T1] is a standard

Brownian motion until time T1 under P2 and under P1,2, consequentlyunder P1,2 we have

L(T, T1, T2) = S(T, T1, T2)

= − 1T2 − T1

+(

1T2 − T1

+ S(t, T1, T2))er Tt

(ζ1s−ζ

2s )dB2

s− 12

r Tt

(ζ1s−ζ

2s )2ds,

has same law as

1T2 − T1

(P (t, T1)P (t, T2)e

X− 12 Var[X] − 1

),

where X is a centered Gaussian random variable with variancew T1

t(ζ1s − ζ2

s )2ds

given Ft. Hence

(T2 − T1) IE[e−


∣∣∣Ft]= P (t, T1, T2)

×Bl(

1T2 − T1

+ S(t, T1, T2),r T1t

(ζ1s − ζ2

s )2ds

T1 − t, κ+ 1

T2 − T1, T1 − t

).

Exercise 12.3

(i) We have

L(t, T1, T2) = L(0, T1, T2)er t

0 γ1(s)dW 2s− 1

2r t

0 |γ1(s)|2ds, 0 ≤ t ≤ T1,

and L(t, T2, T3) = b. Note that we have P (t, T2)/P (t, T3) = 1+δb henceP2 = P3 = P1,2 up to time T1.

(ii) We have

E[e−


∣∣∣Ft]= P (t, T2)E2

[(L(T1, T1, T2)− κ)+ | Ft

]= P (t, T2)E2

[(L(t, T1, T2)e

r T1t γ1(s)dW 2

s− 12

r T1t |γ1(s)|2ds − κ)+ | Ft

]= P (t, T2)Bl(κ, L(t, T1, T2), σ1(t), 0, T1 − t),

whereσ2

1(t) = 1T1 − t

w T1

t|γ1(s)|2ds.

(iii) We have

" 53

N. Privault

P (t, T1)P (t, T1, T3) = P (t, T1)

δP (t, T2) + δP (t, T3)

= P (t, T1)δP (t, T2)

11 + P (t, T3)/P (t, T2)

= 1 + δb

δ(δb+ 2)(1 + δL(t, T1, T2)), 0 ≤ t ≤ T1,

and

P (t, T3)P (t, T1, T3) = P (t, T3)

P (t, T2) + P (t, T3)

= 11 + P (t, T2)/P (t, T3)

= 1δ

12 + δb

, 0 ≤ t ≤ T2. (S.12.18)

(iv) We have

S(t, T1, T3) = P (t, T1)P (t, T1, T3) −

P (t, T3)P (t, T1, T3)

= 1 + δb

δ(2 + δb) (1 + δL(t, T1, T2))− 1δ(2 + δb)

= 12 + δb

(b+ (1 + δb)L(t, T1, T2)), 0 ≤ t ≤ T2.

We have

dS(t, T1, T3) = 1 + δb

2 + δbL(t, T1, T2)γ1(t)dW 2

t

=(S(t, T1, T3)− b

2 + δb

)γ1(t)dW 2

t

= S(t, T1, T3)σ1,3(t)dW 2t , 0 ≤ t ≤ T2,

with

σ1,3(t) =(

1− b

S(t, T1, T3)(2 + δb)

)γ1(t)

=(

1− b

b+ (1 + δb)L(t, T1, T2)

)γ1(t)

= (1 + δb)L(t, T1, T2)b+ (1 + δb)L(t, T1, T2)γ1(t)

= (1 + δb)L(t, T1, T2)(2 + δb)S(t, T1, T3)γ1(t).

(v) The process (W 2)t∈R+ is a standard Brownian motion under P2 and

54

July 22, 2018

"


P (t, T1, T3)E1,3[(S(T1, T1, T3)− κ)+ | Ft

]= P (t, T2)Bl(κ, S(t, T1, T2), σ1,3(t), 0, T1 − t),

where |σ1,3(t)|2 is the approximation of the volatility

1T1 − t

w T1

t|σ1,3(s)|2ds = 1

T1 − t

w T1

t

((1 + δb)L(s, T1, T2)(2 + δb)S(s, T1, T3)

)2γ1(s)ds

obtained by freezing the random component of σ1,3(s) at time t, i.e.

σ21,3(t) = 1

T1 − t

((1 + δb)L(t, T1, T2)(2 + δb)S(t, T1, T3)

)2 w T1

t|γ1(s)|2ds.

Exercise 12.4

1. We have

IE[e−

r Ttrsds (P (T, S)− κ)+

∣∣∣Ft] = VT = V0 +w T

0dVt

= P (0, T ) IET[(P (T, S)− κ)+

]+

w t

0ξTs dP (s, T ) +

w t

0ξSs dP (s, S).

2. We have

dVt = d(e−

r t0 rsdsVt

)= −rte−

r t0 rsdsVtdt+ e−

r t0 rsdsdVt

= −rte−r t

0 rsds(ξTt P (t, T )+ξSt P (t, S))dt+ e−

r t0 rsdsξTt dP (t, T ) + e−

r t0 rsdsξSt dP (t, S)

= ξTt dP (t, T ) + ξSt dP (t, S).

3. By Itô’s formula we have

IET[(P (T, S)− κ)+ |Ft

]= C(XT , 0, 0)

= C(X0, T, v(0, T )) +w t

0

∂C

∂x(Xs, T − s, v(s, T ))dXs

= IET[(P (T, S)− κ)+

]+

w t

0

∂C

∂x(Xs, T − s, v(s, T ))dXs,

since the processt 7→ IET

[(P (T, S)− κ)+ |Ft

]is a martingale under P.

4. We have

dVt = d(Vt/P (t, T ))

" 55

N. Privault

= d IET[(P (T, S)− κ)+ |Ft

]= ∂C

∂x(Xt, T − t, v(t, T ))dXt

= P (t, S)P (t, T )

∂C

∂x(Xt, T − t, v(t, T ))(σSt − σTt )dBTt .

5. We have

dVt = d(P (t, T )Vt)= P (t, T )dVt + VtdP (t, T ) + dVt · dP (t, T )

= P (t, S)∂C∂x

(Xt, T − t, v(t, T ))(σSt − σTt )dBTt + VtdP (t, T )

+P (t, S)∂C∂x

(Xt, T − t, v(t, T ))(σSt − σTt )σTt dt

= P (t, S)∂C∂x

(Xt, T − t, v(t, T ))(σSt − σTt )dBt + VtdP (t, T ).

6. We have

dVt = d(e−r t

0 rsdsVt)= −rte−

r t0 rsdsVtdt+ e−

r t0 rsdsdVt

= P (t, S)∂C∂x

(Xt, T − t, v(t, T ))(σSt − σTt )dBt + VtdP (t, T ).

7. We have

dVt = P (t, S)∂C∂x

(Xt, T − t, v(t, T ))(σSt − σTt )dBt + VtdP (t, T )

= ∂C

∂x(Xt, T − t, v(t, T ))dP (t, S)

−P (t, S)P (t, T )

∂C

∂x(Xt, T − t, v(t, T ))dP (t, T ) + VtdP (t, T )

=(Vt −

P (t, S)P (t, T )

∂C

∂x(Xt, T − t, v(t, T ))

)dP (t, T )

+∂C

∂x(Xt, T − t, v(t, T ))dP (t, S),

hence the hedging strategy (ξTt , ξSt )t∈[0,T ] of the bond option is given by

ξTt = Vt −P (t, S)P (t, T )

∂C

∂x(Xt, T − t, v(t, T ))

= C(Xt, T − t, v(t, T ))− P (t, S)P (t, T )

∂C

∂x(Xt, T − t, v(t, T )),

and

56

July 22, 2018

"


ξSt = ∂C

∂x(Xt, T − t, v(t, T )),

t ∈ [0, T ].8. We have

∂C

∂x(x, τ, v)

= ∂

∂x

[xΦ

(v√τ

2 + 1v√τ

log xκ

)− κΦ

(−v√τ

2 + 1v√τ

log xκ

)]= x

∂

∂xΦ

(v√τ

2 + 1v√τ

log xκ

)− κ ∂

∂xΦ

(−v√τ

2 + 1v√τ

log xκ

)+Φ

(v√τ

2 + 1v√τ

log xκ

)

= xe− 1

2

(v√τ

2 + 1v√τ

log xκ

)2

√2π

(1

v√τx

)− κe

− 12

(− v√τ

2 + 1v√τ

log xκ

)2

√2π

(1

v√τx

)+Φ

(v√τ

2 + 1v√τ

log xκ

)= Φ

(log(x/κ) + τv2/2√

τv

).

As a consequence we get

ξTt = C(Xt, T − t, v(t, T ))− P (t, S)P (t, T )

∂C

∂x(Xt, T − t, v(t, T ))

= P (t, S)P (t, T )Φ

((T − t)v2(t, T )/2 + logXt√

T − tv(t, T )

)−κΦ

(−v(t, T )

2 + 1v(t, T ) log P (t, S)

κP (t, T )

)−P (t, S)P (t, T )Φ

(log(Xt/κ) + (T − t)v2(t, T )/2√

T − tv(t, T )

)= −κΦ

(log(Xt/κ)− (T − t)v2(t, T )/2

v(t, T )√T − t

),

and

ξSt = ∂C

∂x(Xt, T − t, v(t, T )) = Φ

(log(Xt/κ) + (T − t)v2(t, T )/2

v(t, T )√T − t

),

t ∈ [0, T ], and the hedging strategy is given by

VT = IE[e−

r Ttrsds (P (T, S)− κ)+

∣∣∣Ft]

" 57

N. Privault

= V0 +w t

0ξTs dP (s, T ) +

w t

0ξSs dP (s, S)

= V0 − κw t

0Φ

(log(Xt/κ)− (T − t)v2(t, T )/2

v(t, T )√T − t

)dP (s, T )

+w t

0Φ

(log(Xt/κ) + (T − t)v2(t, T )/2

v(t, T )√T − t

)dP (s, S).

Consequently the bond option can be hedged by shortselling a bond withmaturity T for the amount

κΦ

(log(Xt/κ)− (T − t)v2(t, T )/2

v(t, T )√T − t

),

and by buying a bond with maturity S for the amount

Φ

(log(Xt/κ) + (T − t)v2(t, T )/2

v(t, T )√T − t

).

Exercise 12.5

1. We have

S(Ti, Ti, Tj) = S(t, Ti, Tj) exp(w Ti

tσi,j(s)dBi,js −

12

w Ti

t|σi,j |2(s)ds

).

2. We have

P (t, Ti, Tj) IEi,j[(S(Ti, Ti, Tj)− κ)+

∣∣∣Ft]= P (t, Ti, Tj) IEi,j

[(S(t, Ti, Tj)e

r Tit σi,j(s)dBi,js − 1

2r Tit |σi,j |

2(s)ds − κ)+ ∣∣∣Ft]

= P (t, Ti, Tj)Bl(κ, v(t, Ti)/√Ti − t, 0, Ti − t)

= P (t, Ti, Tj)

×(S(t, Ti, Tj)Φ

(log(x/K)v(t, Ti)

+ v(t, Ti)2

)− κΦ

(log(x/K)v(t, Ti)

− v(t, Ti)2

)),

wherev2(t, Ti) =

w Ti

t|σi,j |2(s)ds.

3. Integrate the self-financing condition (12.25) between 0 and t.4. We have

dVt = d(e−

r t0 rsdsVt

)= −rte−

r t0 rsdsVtdt+ e−

r t0 rsdsdVt

58

July 22, 2018

"


= −rte−r t

0 rsds

j∑k=i

ξkt P (t, Tk), dt+ e−r t

0 rsds

j∑k=i

ξkt dP (t, Tk)

=j∑k=i

ξkt dP (t, Tk), 0 ≤ t ≤ Ti.

sincedP (t, Tk)P (t, Tk)

= ζk(t)dt, k = i, . . . , j.

5. We apply the Itô formula and the fact that

t 7→ IEi,j[(S(Ti, Ti, Tj)− κ)+

∣∣∣Ft]and (St)t∈R+ are both martingales under Pi,j .

6. Use the fact that

Vt = IEi,j[(S(Ti, Ti, Tj)− κ)+

∣∣∣Ft] ,and apply the result of Question 5.

7. Apply the Itô rule to Vt = P (t, Ti, Tj)Vt using Relation (12.23) and theresult of Question 6.

8. We have

dVt = St∂C

∂x(St, v(t, Ti))

×

(j−1∑k=i

(Tk+1 − Tk)P (t, Tk+1)(ζi(t)− ζk+1(t)) + P (t, Tj)(ζi(t)− ζj(t)))dBt

+VtdP (t, Ti, Tj)

= St∂C

∂x(St, v(t, Ti))

×

(j−1∑k=i

(Tk+1 − Tk)P (t, Tk+1)(ζi(t)− ζk+1(t)) + P (t, Tj)(ζi(t)− ζj(t)))dBt

+Vtj−1∑k=i

(Tk+1 − Tk)ζk+1(t)P (t, Tk+1)dBt

= St∂C

∂x(St, v(t, Ti))

j−1∑k=i

(Tk+1 − Tk)P (t, Tk+1)(ζi(t)− ζk+1(t))dBt

+∂C

∂x(St, v(t, Ti))P (t, Tj)(ζi(t)− ζj(t))dBt

+Vtj−1∑k=i


" 59

N. Privault

= Stζi(t)∂C

∂x(St, v(t, Ti))

j−1∑k=i

(Tk+1 − Tk)P (t, Tk+1)dBt

−St∂C

∂x(St, v(t, Ti))

j−1∑k=i

(Tk+1 − Tk)P (t, Tk+1)ζk+1(t)dBt

+∂C


+Vtj−1∑k=i


= Stζi(t)∂C

∂x(St, v(t, Ti))

j−1∑k=i


+(Vt − St

∂C

∂x(St, v(t, Ti))

) j−1∑k=i


+∂C

∂x(St, v(t, Ti))P (t, Tj)(ζi(t)− ζj(t))dBt.

9. We have

dVt = Stζi(t)∂C

∂x(St, v(t, Ti))

j−1∑k=i


+(Vt − St

∂C

∂x(St, v(t, Ti))

) j−1∑k=i


+∂C


= (P (t, Ti)− P (t, Tj))ζi(t)∂C

∂x(St, v(t, Ti))dBt

+(Vt − St

∂C

∂x(St, v(t, Ti))

)dP (t, Ti, Tj)

+∂C


= (ζi(t)P (t, Ti)− ζj(t)P (t, Tj))∂C

∂x(St, v(t, Ti))dBt

+(Vt − St

∂C

∂x(St, v(t, Ti))

)dP (t, Ti, Tj)

= ∂C

∂x(St, v(t, Ti))d(P (t, Ti)− P (t, Tj))

+(Vt − St

∂C

∂x(St, v(t, Ti))

)dP (t, Ti, Tj).

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10. We have

∂C

∂x(x, τ, v) = ∂

∂x

[xΦ

(v

2 + 1v

log xκ

)− κΦ

(−v2 + 1

vlog x

κ

)]= x

∂

∂xΦ

(v

2 + 1v

log xκ

)− κ ∂

∂xΦ

(−v2 + 1

vlog x

κ

)+ Φ

(v

2 + 1v

log xκ

)= x

e−12 ( v2 + 1

v log xκ )2

√2π

(1vx

)− κe

− 12 (− v2 + 1

v log xκ )2

√2π

(1vx

)+Φ

(v

2 + 1v

log xκ

)= Φ

(log(x/κ)

v+ v

2

).

11. We have

dVt = ∂C

∂x(St, v(t, Ti))d(P (t, Ti)− P (t, Tj))

+(Vt − St

∂C

∂x(St, v(t, Ti))

)dP (t, Ti, Tj)

= Φ

(log(St/K)v(t, Ti)

+ v(t, Ti)2

)d(P (t, Ti)− P (t, Tj))

−κΦ(

log(St/K)v(t, Ti)

− v(t, Ti)2

)dP (t, Ti, Tj).

12. We compare the results of Questions 4 and 11.

Chapter 13

Exercise 13.1 Defaultable bonds.

1. Use the fact that (rt, λt)t∈[0,T ] is a Markov process.2. Use the “tower property” (16.22) for the conditional expectation givenFt.

3. We have

d(e−

r t0 (rs+λs)dsP (t, T )

)= −(rt + λt)e−

r t0 (rs+λs)dsP (t, T )dt+ e−

r t0 (rs+λs)dsdP (t, T )

= −(rt + λt)e−r t

0 (rs+λs)dsP (t, T )dt+ e−r t

0 (rs+λs)dsdF (t, rt, λt)

= −(rt + λt)e−r t

0 (rs+λs)dsP (t, T )dt+ e−r t

0 (rs+λs)ds ∂F

∂x(t, rt, λt)drt

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N. Privault

+e−r t

0 (rs+λs)ds ∂F

∂y(t, rt, λt)dλt + 1

2e−

r t0 (rs+λs)ds ∂

2F

∂x2 (t, rt, λt)σ21(t, rt)dt

+12e−

r t0 (rs+λs)ds ∂

2F

∂y2 (t, rt, λt)σ22(t, λt)dt

+e−r t

0 (rs+λs)dsρ∂2F

∂x∂y(t, rt, λt)σ1(t, rt)σ2(t, λt)dt+ e−

r t0 (rs+λs)ds ∂F

∂t(t, rt, λt)dt

= e−r t

0 (rs+λs)ds ∂F

∂x(t, rt, λt)σ1(t, rt)dB(1)

t + e−r t

0 (rs+λs)ds ∂F

∂y(t, rt, λt)σ2(t, λt)dB(2)

t

+e−r t

0 (rs+λs)ds(−(rt + λt)P (t, T ) + ∂F

∂x(t, rt, λt)µ1(t, rt)

+∂F

∂y(t, rt, λt)µ2(t, λt) + 1

2∂2F

∂x2 (t, rt, λt)σ21(t, rt) + 1

2∂2F

∂y2 (t, rt, λt)σ22(t, λt)

+ρ ∂2F

∂x∂y(t, rt, λt)σ1(t, rt)σ2(t, λt) + ∂F

∂t(t, rt, λt)

)dt,

hence the bond pricing PDE is

−(x+ y)F (t, x, y) + µ1(t, x)∂F∂x

(t, x, y)

+µ2(t, y)∂F∂y

(t, x, y) + 12σ

21(t, x)∂

2F

∂x2 (t, x, y)

+12σ

22(t, y)∂

2F

∂y2 (t, x, y) + ρσ1(t, x)σ2(t, y) ∂2F

∂x∂y(t, x, y) + ∂F

∂t(t, rt, λt) = 0.

4. We havew t

0rsds = 1

a

(σB

(1)t − rt

)= σ

a

(B

(1)t −

w t

0e−a(t−s)dB(1)

s

)= σ

a

w t

0(1− e−a(t−s))dB(1)

s ,

hencew T

trsds =

w T

0rsds−

w t

0rsds

= σ

a

w T

0(1− e−a(T−s))dB(1)

s −σ

a

w t

0(1− e−a(t−s))dB(1)

s

= −σa

(w t

0(e−a(T−s) − e−a(t−s))dB(1)

s +w T

t(e−a(T−s) − 1)dB(1)

s

)= −σ

a(e−a(T−t) − 1)

w t

0e−a(t−s)dB(1)

s −σ

a

w T

t(e−a(T−s) − 1)dB(1)

s

= −1a

(e−a(T−t) − 1)rt −σ

a

w T

t(e−a(T−s) − 1)dB(1)

s .

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The answer for λt is similar.5. As a consequence of the previous question we have

IE[w T

trsds+

w T

tλsds

∣∣∣Ft] = C(a, t, T )rt + C(b, t, T )λt,

and

Var[w T

trsds+

w T

tλsds

∣∣∣Ft] =

= Var[w T

trsds

∣∣∣Ft]+ Var[w T

tλsds

∣∣∣Ft]+2Cov

(w T

tXsds,

w T

tYsds

∣∣∣Ft)= σ2

a2

w T

t(e−a(T−s) − 1)2ds

+2ρσηab

w T

t(e−a(T−s) − 1)(e−b(T−s) − 1)ds

+η2

b2

w T

t(e−b(T−s) − 1)2ds

= σ2w T

tC2(a, s, T )ds+ 2ρση

w T

tC(a, s, T )C(b, s, T )ds

+η2w T

tC2(b, sT )ds,

from the Itô isometry.6. We have

P (t, T ) = 1τ>t IE[exp

(−

w T

trsds−

w T

tλsds

) ∣∣∣Ft]= 1τ>t exp

(− IE

[w T

trsds

∣∣∣Ft]− IE[w T

tλsds

∣∣∣Ft])× exp

(12 Var

[w T

trsds+

w T

tλsds

∣∣∣Ft])= 1τ>t exp (−C(a, t, T )rt − C(b, t, T )λt)

× exp(σ2

2

w T

tC2(a, s, T )ds+ η2

2

w T

tC2(b, s, T )e−b(T−s)ds

)× exp

(ρση

w T


).

7. This is a direct consequence of the answers to Questions 3 and 6.8. The above analysis shows that

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N. Privault

P(τ > T | Gt) = 1τ>t IE[exp

(−

w T

tλsds

) ∣∣∣Ft]= 1τ>t exp

(−C(b, t, T )λt + η2

2

w T

tC2(b, s, T )ds

),

for a = 0 and

IE[exp

(−

w T

trsds

) ∣∣∣Ft] = exp(−C(a, t, T )rt + σ2

2

w T

tC2(a, s, T )ds

),

for b = 0, and this implies

U(t, T ) = exp(ρση

w T


)= exp

(ρση

ab(T − t− C(a, t, T )− C(b, t, T ) + C(a+ b, t, T ))

).

9. We have

f(t, T ) = −1τ>t∂ logP (t, T )

∂T

= 1τ>t(rte−a(T−t) − σ2

2 C2(a, t, T ) + λte−b(T−t) − η2

2 C2(b, t, T )

)−1τ>tρσηC(a, t, T )C(b, t, T ).

10. We use the relation

P(τ > T | Gt) = 1τ>t IE[exp

(−

w T

tλsds

) ∣∣∣Ft]= 1τ>t exp

(−C(b, t, T )λt + η2

2

w T

tC2(b, s, T )ds

)= 1τ>te−

r Ttf2(t,u)du,

where f2(t, T ) is the Vasiçek forward rate corresponding to λt, i.e.

f2(t, u) = λte−b(u−t) − η2

2 C2(b, t, u).

11. In this case we have ρ = 0 and

P (t, T ) = P(τ > T | Gt) IE[exp

(−

w T

trsds

) ∣∣∣Ft] ,since U(t, T ) = 0.

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Chapter 14

Exercise 14.1

1. We have St = S0e−ληt(1 + η)Nt , t ∈ R+.

2. We have St = S0ert

Nt∏k=1

(1 + ηZk), t ∈ R+.


Var[YT ] = E

(NT∑k=1

Zk − E[YT ])2

=∞∑n=0

E

(NT∑k=1

Zk − λtE[Z1])2 ∣∣∣NT = k

P(NT = k)

= e−λt∞∑n=0

λntn

n! E

( n∑k=1

Zk − λtE[Z1])2

= e−λt∞∑n=0

λntn

n! E

( n∑k=1

Zk

)2

− 2λtE[Z1]n∑k=1

Zk + λ2t2(E[Z1])2

= e−λt

∞∑n=0

λntn

n!

×E

2∑

1≤k<l≤nZkZl +

n∑k=1|Zk|2 − 2λtE[Z1]

n∑k=1

Zk + λ2t2(E[Z1])2

= e−λt

∞∑n=0

λntn

n!

×(n(n− 1)(E[Z1])2 + nE[|Z1|2]− 2nλt(E[Z1])2 + λ2t2(E[Z1])2)

= e−λt(E[Z1])2∞∑n=2

λntn

(n− 2)! + e−λtE[|Z1|2]∞∑n=1

λntn

(n− 1)!

−2e−λtλt(E[Z1])2∞∑n=1

λntn

(n− 1)! + λ2t2(E[Z1])2)

= λtE[|Z1|2],

or, using the characteristic function of Proposition 14.3,

Var[YT ] = E[|YT |2]− (E[YT ])2

= − d2

dα2E[eiαYT ]|α=0 − λ2t2(E[Z1])2

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N. Privault

= λtw∞

−∞|y|2µ(dy) = λtE[|Z1|2].

Exercise 14.3

1. We have

dSt =(µ+ 1

2σ2)Stdt+ σStdWt + (St − St−)dNt

=(µ+ 1

2σ2)Stdt+ σStdWt + (S0e

µt+σWt+Yt − S0eµt+σWt+Yt−)dNt

=(µ+ 1

2σ2)Stdt+ σStdWt + (S0e

µt+σWt+Yt−+ZNt − eµt+σWt+Yt−)dNt

=(µ+ 1

2σ2)Stdt+ σStdWt + St−(eZNt − 1)dNt,

hence the jumps of St are given by the sequence (eZk − 1)k≥1.2. The discounted process e−rtSt satisfies

d(e−rtSt) = e−rt(µ− r + 1

2σ2)Stdt+σe−rtStdWt+e−rtSt−(eZNt−1)dNt.

Hence by the Girsanov theorem, choosing u, λ, ν such that

µ− r + 12σ

2 = u− λ IEν [eZ1 − 1],

shows that

d(e−rtSt) = σe−rtSt(dWt+udt)+e−rtSt−((eZNt−1)dNt−λ IEν [eZ1−1]dt)

is a martingale under (Pu,λ,ν).

Exercise 14.4

1. We have

St = S0eµt

Nt∏k=1

(1 + Yk) = S0 exp(µt+

Nt∑k=1

Xk

), t ∈ R+.

2. We have

e−rtSt = S0 exp(

(µ− r)t+Nt∑k=1

Xk

), t ∈ R+,

which is a martingale if

0 = µ− r + λ IE[Yk] = µ− r + λ IE[eXk − 1] = µ− r + λ(eσ2/2 − 1).

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3. We have

e−r(T−t) IE[(ST − κ)+ | St]

= e−r(T−t) IE

(S0 exp(µT +

Nt∑k=1

Xk

)− κ

)+ ∣∣∣St

= e−r(T−t)∞∑n=0

IE[(Ste

µ(T−t)+∑n

k=1Xk − κ

)+ ∣∣∣St]P(NT −Nt = n)

= e−(r+λ)(T−t)∞∑n=0

IE[(Ste

µ(T−t)+∑n

k=1Xk − κ

)+ ∣∣∣St] (λ(T − t))n

n!

= e−λ(T−t)∞∑n=0

Bl(Ste(µ−r)(T−t)+nσ2/2, r, nσ2/(T − t), κ, T − t) (λ(T − t))n

n!

= e−λ(T−t)∞∑n=0

(Ste

(µ−r)(T−t)+nσ2/2Φ(d+)− κe−r(T−t)Φ(d−)) (λ(T − t))n

n! ,

with

d+ = log(Ste(µ−r)(T−t)+nσ2/2/κ) + r(T − t) + nσ2/2√nσ

= log(St/κ) + µ(T − t) + nσ2√nσ

,

d− = log(Ste(µ−r)(T−t)+nσ2/2/κ) + r(T − t)− nσ2/2√nσ

= log(St/κ) + µ(T − t)√nσ

,

and µ = r + λ(1− eσ2/2).

Chapter 15

Exercise 15.1

1. Independently of the choice of a risk-neutral measure Pu,λ,ν we have

e−r(T−t) IEu,λ,ν [ST −K | Ft] = ert IEu,λ,ν [e−rTST | Ft]−Ke−r(T−t)

= erte−rtSt −Ke−r(T−t)

= St −Ke−r(T−t)

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N. Privault

= f(t, St),

forf(t, x) = x−Ke−r(T−t), t, x > 0.

2. Clearly, holding one unit of the risky asset and shorting a (possibly frac-tional) quantity Ke−rT of the riskless asset will hedge the payoff ST −K,and this hedging strategy is self-financing because it is constant in time.

3. Since ∂f∂x

(t, x) = 1 we have

ξt =σ2 ∂f

∂x(t, St−) + aλ

St−(f(t, St−(1 + a))− f(t, St−))

σ2 + a2λ

=σ2 + aλ

St−(St−(1 + a)− St−)

σ2 + a2λ= 1, t ∈ [0, T ],

which coincides with the result of Question 2.

Exercise 15.2

(i) We have

St = S0 exp(µt+ σBt −

12σ

2t

)(1 + η)Nt .

(ii) We have

St = S0 exp(

(µ− r)t+ σBt −12σ

2t

)(1 + η)Nt ,

anddSt = (µ− r + λη)Stdt+ ηSt−(dNt − λdt) + σStdWt,

hence we need to takeµ− r + λη = 0,

since the compensated Poisson process (Nt − λt)t∈R+ is a martingale.(iii) We have

e−r(T−t)E∗[(ST − κ)+ | St]

= e−r(T−t)E∗[(

S0 exp(µT + σBT −

12σ

2T

)(1 + η)NT − κ

)+ ∣∣∣St]

= e−r(T−t)E∗[(Ste

µ(T−t)+σ(BT−Bt)− 12σ

2(T−t)(1 + η)NT−Nt − κ)+ ∣∣∣St]

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July 22, 2018

"


= e−r(T−t)∞∑n=0

P(NT −Nt = n)

×E∗[(Ste

µ(T−t)+σ(BT−Bt)− 12σ

2(T−t)(1 + η)n − κ)+ ∣∣∣St]

= e−(r+λ)(T−t)∞∑n=0

(λ(T − t))n

n!

×E∗[(Ste

(r−λη)(T−t)+σ(BT−Bt)− 12σ

2(T−t)(1 + η)n − κ)+ ∣∣∣St]

= e−λ(T−t)∞∑n=0

Bl(Ste−λη(T−t)(1 + η)n, r, σ2, T − t, κ) (λ(T − t))n

n!

= e−λ(T−t)∞∑n=0

(Ste−λη(T−t)(1 + η)nΦ(d+)− κe−r(T−t)Φ(d−)

) (λ(T − t))n

n! ,

with

d+ = log(Ste−λη(T−t)(1 + η)n/κ) + (r + σ2/2)(T − t)σ√T − t

= log(St(1 + η)n/κ) + (r − λη + σ2/2)(T − t)σ√T − t

,

and

d− = log(Ste−λη(T−t)(1 + η)n/κ) + (r − σ2/2)(T − t)σ√T − t

= log(St(1 + η)n/κ) + (r − λη − σ2/2)(T − t)σ√T − t

.

Exercise 15.3

1. The discounted process St = e−rtSt satisfies the equation

dSt = YNt St−dNt, ,

and it is a martingale since the compound Poisson process YNtdNt iscentered with independent increments as IE[Y1] = 0.

2. We have

ST = S0erT

NT∏k=1

(1 + Yk),

hence

e−rT IE[(ST − κ)] = e−rT IE

(S0erT

NT∏k=1

(1 + Yk)− κ)+

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N. Privault

= e−rT∞∑n=0

IE

(S0erT

NT∏k=1

(1 + Yk)− κ)+ ∣∣∣NT = n

P(NT = n)

= e−rT−λT∞∑k=0

IE

(S0erT

n∏k=1

(1 + Yk)− κ)+ (λT )n

n!

= e−rT−λT∞∑k=0

(λT )n

2nn!

w 1

−1· · ·

w 1

−1

(S0e

rTn∏k=1

(1 + yk)− κ)+

dy1 · · · dyn.

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